ICERM School Notes: Complete Segal Spaces (v0.3) · sHoTT / Rzk · RS17

Simplicial Sets & Complete Segal Spaces: A Visual and Interactive Library

This companion tool was created using Claude Opus 5 and Claude Fable 5. Its content has been checked, but errors may remain — please write to uskupluelif@gmail.com.

Section 2.1 · Notes: Def. 2.1 – Lemma 2.5The simplex category ∆, chain notation, and unique decomposition

Definition 2.1 · Simplex category The category ∆ is defined as follows: the objects are the linearly ordered sets [n] := {0 ≤ 1 ≤ ⋯ ≤ n} for n ≥ 0; Hom([m], [n]) is the set of all non-decreasing maps from [m] to [n].
Notation 2.2 · Chain notation A morphism δ : [m] → [n] is precisely a non-decreasing chain 0 ≤ a₀ ≤ a₁ ≤ ⋯ ≤ am ≤ n, where aᵢ is the image of i ∈ [m]. We denote such a morphism by a₀a₁...am. For example, the chain 02 is the morphism [1] → [2] sending 0 ↦ 0 and 1 ↦ 2.
Notation 2.3 · Generating morphisms dⁱ and sⁱ

• For each n ≥ 0 and 0 ≤ i ≤ n+1 there is a unique injective map dⁱ : [n] → [n+1] whose image does not contain i (coface).

• For each n ≥ 1 and 0 ≤ i ≤ n−1 there is a unique surjective map sⁱ : [n] → [n−1] whose pre-image over i is {i, i+1} and whose other pre-images have one element (codegeneracy).

Tool · Write dⁱ and sⁱ as chains

Pick n and i; see the chain notation and element-by-element behavior of dⁱ : [n] → [n+1] and sⁱ : [n] → [n−1].

Pick n and i, then press "Show".
Lemma 2.4 · Unique decomposition An arbitrary morphism δ : [m] → [n] in ∆ can be uniquely written as a composition of the form
δ = di₁ di₂ ⋯ dik sj₁ ⋯ sjl
where 0 ≤ ik < ⋯ < i₂ < i₁ ≤ n and 0 ≤ j₁ < j₂ < ⋯ < jl < m.

Idea: the s's first "compress" the chain (repeats are collapsed — the surjective part), then the d's "skip" the elements missing from the image (the injective part). The d exponents are the elements of [n] not in the image (largest first); the s exponents are the positions j with aj = aj+1 (smallest first).

Tool · Unique decomposition calculator

Enter the target n and the chain (e.g. n = 3 and chain 022, that is 022 : [2] → [3]). The tool produces the decomposition of Lemma 2.4 together with a verification.

Enter a chain and press "Decompose".
Exercise 2.1

Write down explicitly the maps d⁰, d¹, d² : [1] → [2] and s⁰ : [1] → [0] using the notation from Notation 2.2.

Show solution
dⁱ is the injective map whose image does not contain i: d⁰ = 12, d¹ = 02, d² = 01. s⁰ : [1] → [0] sends everything to 0: s⁰ = 00.
Exercise 2.16

Express the maps 001, 011 ∈ Hom([2], [1]) in terms of the generating morphisms sⁱ (write your answer in the form s0).

Show solution
In 001 the repeat is at position 0 (a₀ = a₁), so 001 = s⁰. In 011 the repeat is at position 1, so 011 = s¹. (You can verify with the calculator above using n = 1.)

Section 2.1 (cont.) · Notes: Def. 2.7 – Rem. 2.11Simplicial sets and the maps dᵢ, sᵢ

Definition 2.7 / 2.8 · Simplicial set Let C be a category. A simplicial object in C is a functor X : ∆op → C. A simplicial set is a simplicial object in Set. The category of simplicial sets is sSet := Fun(∆op, Set).

Notation 2.9: For δ : [m] → [n], the corresponding map X(δ) : Xn → Xm is denoted δ. In particular, for dⁱ and sⁱ the corresponding maps are written dᵢ = (dⁱ) and sᵢ = (sⁱ). Note: the generators of ∆ carry upper indices (dⁱ, sⁱ), the maps in a simplicial set carry lower indices (dᵢ, sᵢ), and all arrows are reversed: dᵢ : Xn+1 → Xn, sᵢ : Xn−1 → Xn.

Intuition 2.13 (geometric intuition): We imagine X as a recipe for constructing a topological space with a notion of "direction": X₀ is the set of vertices; the non-degenerate elements of X₁ are the edges; the non-degenerate elements of X₂ are the triangles, and so on. The maps d₀, d₁ : X₁ → X₀ specify the source and target of each edge, while d₀, d₁, d₂ : X₂ → X₁ specify the edges forming the boundary of each triangle. Degenerate simplices do not manifest geometrically, but they play an important role (see Section 6, Remark 2.46).

Section 2.2 · Notes: Def. 2.16 – Cor. 2.27The standard simplex Dn: how face and degeneracy maps behave

Warning 0.1 · Notation warning The notes denote the standard n-simplex by Dn (instead of the more common ∆n); its boundary by ∂Dn, the k-th horn by Lnk, and the spine by Spn. This choice is made to match the syntax of Rzk (see Section 7).
Definition 2.16 · Standard simplex For a given n ≥ 0, the standard n-simplex is the representable functor Dn := Hom(−, [n]). That is, Dnm = Hom([m], [n]): the m-simplices of Dn are the chains [m] → [n].
Remark 2.17 · dᵢ and sᵢ on chains

dᵢ : Dnm+1 → Dnm deletes the i-th element of the chain.

sᵢ : Dnm−1 → Dnm repeats the i-th element of the chain.

Remark 2.18: Since sᵢ adds repeated elements, an m-simplex a₀...am is non-degenerate ⇔ the chain has no repeated elements (the map is injective).

D⁰, D¹, D² geometrically (Ex. 2.19, 2.22, 2.24 and Intuition 2.21, 2.23, 2.25)

0
D⁰ — single non-degenerate simplex: 0. A single point.
0 1 01
D¹ — non-degenerate: 0, 1, 01. A single directed edge.
0 1 2 01 12 02 012
D² — seven non-degenerate simplices: 0, 1, 2; 01, 12, 02; 012.

The levels (Example 2.24): D²₀ = {0, 1, 2}, D²₁ = {00, 01, 11, 12, 22, 02}, D²₂ = {000, 001, 011, 111, 112, 122, 222, 002, 022, 012}, ...

Tool · D² explorer: dᵢ and sᵢ on chains and on the picture

Pick a simplex, then apply dᵢ (delete) or sᵢ (repeat). On the right, the geometric carrier of the chain (its image set) is highlighted on the triangle; degenerate simplices "collapse" onto a lower-dimensional carrier.

Pick from D²₁:
Pick from D²₂:
Pick a chain.
0 1 2
Tool · The same maps one dimension up: D³ explorer in 3D

Drag the tetrahedron to rotate it. Pick a simplex from the lists (chains with a repeat are marked in amber — the degenerate ones), or click a vertex, edge, or face directly on the tetrahedron. Then apply dᵢ (delete) or sᵢ (repeat) as above. The geometric carrier is highlighted; degenerate simplices collapse onto a lower-dimensional carrier, and a shaded interior means the carrier is the solid tetrahedron.

Pick from D³₁:
Pick from D³₂:
Pick from D³₃ (35 chains — the unique non-degenerate one is 0123):
Pick a chain, or click on the tetrahedron.
drag to rotate · click to select
Exercise 2.17

Compute the following face and degeneracy maps in D²:


Show solution
dᵢ deletes the i-th element: d₀(012) = 12, d₁(012) = 02, d₂(012) = 01. sᵢ repeats the i-th element: s₀(02) = 002, s₁(02) = 022. You can verify with the explorer above by selecting 012 and using the buttons.
Lemma 2.26 · Yoneda lemma for simplicial sets For a given n ≥ 0 and a simplicial set X, there is a natural isomorphism HomsSet(Dn, X) ≅ Xn. That is, the n-simplices of X are the same thing as maps Dn → X. Corollary 2.27: HomsSet(Dn, Dk) ≅ Hom([n], [k]).

Notes: Def. 2.12, Rem. 2.18Degenerate and non-degenerate simplices

Definition 2.12 · Degenerate simplex Let X be a simplicial set. An n-simplex σ in Xn is called degenerate if there exists a simplex τ in Xn−1 and 0 ≤ i ≤ n−1 such that σ = sᵢ(τ). Otherwise, σ is called non-degenerate.

Visual explanation: In Dn, if a chain contains a repeat, its geometric carrier collapses onto a lower-dimensional simplex. For example, 001 ∈ D²₂ is a "2-simplex", but since it equals s₀(01) it actually lives flattened on the edge 01; it carries no new geometry. In contrast, 012 is injective — it is the genuine triangle face.

01 01 s₀(01) = 001
001 is degenerate: it lives on the edge 01; it is not a triangle.
Tool · Degenerate? Write it as sᵢ(τ)

Enter n and a chain for a simplex of Dn. The tool decides whether the simplex is degenerate; if it is, it gives the expression σ = sᵢ(τ) (with the smallest i) and the full degeneracy decomposition.

Enter a chain and press "Inspect".
Exercise 2.2

For each of the following simplices in D²₂, decide whether it is degenerate or non-degenerate:

Show solution
Remark 2.18: in Dn, a chain with a repeat is degenerate. 001 = s₀(01) is degenerate; 011 = s₁(01) is degenerate; 012 is injective, non-degenerate; 022 = s₁(02) is degenerate.

Section 2.3 · Notes: Def. 2.30 – Ex. 2.44Sub-simplicial sets: boundary ∂Dn, horn Lnᵢ, spine Spn

Definition 2.30 / Remark 2.31 · Sub-simplicial set For a simplicial set X, a sub-simplicial set Y consists of subsets Yn ⊆ Xn for every n such that for every δ : [m] → [n], the map X(δ) : Xn → Xm restricts to a map Y(δ) : Yn → Ym. (That is: the subsets are closed under all face and degeneracy maps.)
Definition 2.33 · Boundary For every n ≥ 0, the boundary of Dn, denoted ∂Dn, is the sub-simplicial set with ∂Dnm := {δ : [m] → [n] | Im(δ) ≠ [n]}. (The non-surjective chains; non-surjectivity is closed under precomposition.)

Example 2.34: ∂D⁰ = ∅. Example 2.35: ∂D¹ consists of the two vertices {0, 1} and their degenerate simplices (there is no edge 01). Example 2.36: ∂D² is the vertices and the three edges of the triangle; the interior face 012 is absent.

Definition 2.37 · Horn Let n ≥ 1 and 0 ≤ i ≤ n. The i-th horn of Dn, denoted Lnᵢ, is the sub-simplicial set with (Lnᵢ)m := {δ : [m] → [n] | Im(δ) ∪ {i} ≠ [n]}. From the definition, Lnᵢ ⊆ ∂Dn: the horn is obtained from the boundary by also removing the face opposite the i-th vertex.

Example 2.38: L¹₀ ≅ L¹₁ ≅ D⁰ (a single vertex). Example 2.39: L²₀ = {edges 01, 02}, L²₁ = {edges 01, 12}, L²₂ = {edges 12, 02} (together with the vertices).

Definition 2.40 · Spine For every n ≥ 2, the spine of Dn, denoted Spn, is the sub-simplicial set with Spnm := {δ : [m] → [n] | δ(m) − δ(0) ≤ 1}. That is, the chain of edges joining consecutive vertices: 0 → 1 → 2 → ⋯ → n.

Lemma 2.41: Spn₀ = Dn₀ (all vertices). Lemma 2.42: for k ≥ 2, every element of Spnk is degenerate (the spine is 1-dimensional). Example 2.43: Sp² ≅ L²₁. Example 2.44: Sp³₁ = {00, 01, 11, 12, 22, 23, 33}.

Tool · Sub-shape selector inside D²

Pick a sub-simplicial set; see which non-degenerate simplices of the triangle remain and which are removed. The first three levels are listed below.

0 1 2
Tool · Sub-shape selector inside D³, in 3D

Drag to rotate. The same sub-shapes, one dimension up. Filled faces are present as 2-simplices; an unfilled face is a hole (in each horn L³ₖ, exactly the face opposite k); faded dashed edges are removed. A shaded solid interior means the 3-simplex 0123 is present — this is what distinguishes D³ from ∂D³. The panel lists the non-degenerate simplices, what is missing, and the chain counts, all computed from the defining condition.

drag to rotate
Tool · Membership checker: which sub-shapes contain a given chain?

Enter a chain [m] → [2] (a simplex of D²). The tool tells you whether the chain belongs to ∂D², L²₀, L²₁, L²₂, and Sp², together with the condition from each definition.

Enter a chain and press "Check".
Tool · Membership checker for D³

The same check for chains [m] → [3]: membership in ∂D³, the four horns, and Sp³, with the condition from each definition. Try 0123, 013, 00112.

How Sp³ sits inside D³ (Example 2.44)

0 1 2 3 01 12 23 02 13 03
Highlighted edges are the non-degenerate 1-simplices of Sp³ (01, 12, 23); the faded edges are the rest of the 1-skeleton of D³ (02, 13, 03). Sp³ sits inside D³ as a sub-simplicial set.
Exercise 2.18

Compute the non-degenerate simplices of ∂D³: how many are there in dimensions 0, 1, 2, and 3?

Show solution
The non-degenerate m-simplices of ∂D³ are the injective but non-surjective chains [m] → [3]. Dimension 0: 4 (vertices 0, 1, 2, 3). Dimension 1: 6 (01, 02, 03, 12, 13, 23). Dimension 2: 4 (012, 013, 023, 123). Dimension 3: 0 (the only injective chain 0123 is surjective, so it is not in ∂D³).
Exercise 2.19

Based on computing (L²₀)₁, (L²₁)₁, (L²₂)₁: for each horn, which non-degenerate edge of ∂D² is missing?

Show solution
The definition of Lnᵢ requires Im(δ) ∪ {i} ≠ [n]; the removed edge is the one whose image together with i fills all of [2], i.e. the edge not containing i: 12 is missing in L²₀, 02 in L²₁, 01 in L²₂. (Compare with the sub-shape selector above.)
Exercise 2.20

Compute the non-degenerate simplices of Sp⁴: list its 0-simplices and 1-simplices (comma-separated). Why are there no non-degenerate simplices in dimension 2 or higher?

Show solution
Lemma 2.41: Sp⁴₀ = D⁴₀ = {0, 1, 2, 3, 4}. The non-degenerate 1-simplices are the injective edges satisfying δ(1) − δ(0) ≤ 1: 01, 12, 23, 34. Lemma 2.42: for k ≥ 2, every element of Sp⁴k is degenerate; an injective chain [k] → [4] with k ≥ 2 must have δ(k) − δ(0) ≥ 2.

Section 2.4 · Notes: Ex. 2.45 – Prop. 2.50How the square is obtained: product and coproduct constructions

Limits and colimits in sSet are computed level-wise: for example (X × Y)n = Xn × Yn and (X ⨿ Y)n = Xn ⨿ Yn, with face and degeneracy maps acting componentwise.

Coproduct: ∂D¹ ≅ D⁰ ⨿ D⁰ (Lemma 2.47)

A map D⁰ ⨿ D⁰ → ∂D¹ amounts, by Yoneda, to picking two elements of ∂D¹₀ = {0, 1}; the map 0 ⨿ 1 is at every level the bijection {00...0} ⨿ {00...0} → {00...0, 11...1}. So two disjoint points = the boundary of the edge.

0 ⨿ 0 0 1 ∂D¹ (no edge)
The coproduct is the level-wise disjoint union.

Pushout: horns are two edges glued together (Lemma 2.48)

Lemma 2.48 There are three isomorphisms:   L²₀ ≅ D¹ ⨿0,0D⁰ D¹  ·  L²₁ ≅ D¹ ⨿1,0D⁰ D¹  ·  L²₂ ≅ D¹ ⨿1,1D⁰ D¹.

The superscripts record at which vertex (0 = source, 1 = target) D⁰ is glued into each copy of D¹. For instance, for L²₁: the target (1) of the first edge is identified with the source (0) of the second.

0 1 2 01 02
L²₀ ≅ D¹ ⨿0,0D⁰ D¹: two edges glued at their common source (0).
0 1 2 01 12
L²₁ ≅ D¹ ⨿1,0D⁰ D¹: target of the first = source of the second (vertex 1).
0 1 2 12 02
L²₂ ≅ D¹ ⨿1,1D⁰ D¹: two edges glued at their common target (2).

Lemma 2.49: By the same idea, Spn ≅ D¹ ⨿D⁰ ⋯ ⨿D⁰ D¹ (n edges glued end to end; Exercise 2.11 by induction).

Product: the square D¹ × D¹ (Example 2.45, Remark 2.46)

The product is also level-wise: (D¹ × D¹)n = D¹n × D¹n; an n-simplex is a pair of chains (α, β). The first three levels:

  • (D¹ × D¹)₀ = {(0,0), (0,1), (1,0), (1,1)} — 4 vertices
  • (D¹ × D¹)₁ — 9 elements; the non-degenerate ones: (00,01), (11,01), (01,00), (01,11), (01,01) — 4 sides + the diagonal
  • (D¹ × D¹)₂ — 16 elements; the non-degenerate ones: (011,001) and (001,011) — two triangles
(0,0) (0,1) (1,0) (1,1) (00,01) (01,00) (01,11) (11,01) (01,01) (011,001) (001,011)
The square = two triangles glued along the diagonal (01,01). The bottom-left triangle is (011,001), the top-right one is (001,011) (layout as in the notes).
Remark 2.46 · The role of degenerate simplices The pair (00, 01) is non-degenerate in D¹ × D¹, even though its first component 00 ∈ D¹₁ is degenerate on its own. A notion of product compatible with our geometric intuition requires the presence of degenerate simplices: the top side (00,01) stays constant in the first coordinate while traversing 01 in the second.
Tool · D¹ × D¹ simplex inspector

Enter two chains of the same length: α, β ∈ D¹n (digits 0 and 1 only). The tool decides whether the pair (α, β) is degenerate and draws its vertex sequence ((α₀,β₀), ..., (αnn)) as a path in the square. A pair is (α, β) = sᵢ(τ) only when both components repeat at the same position i — this explains the phenomenon of Remark 2.46.

Enter two chains.
(0,0) (0,1) (1,0) (1,1)
Exercise 2.22

Identify the non-degenerate 1-simplices and 2-simplices of D¹ × D¹:

1-simplex candidates:
2-simplex candidates:
Show solution
A pair is degenerate ⇔ both components repeat at the same position. Non-degenerate 1-simplices: (00,01), (11,01), (01,00), (01,11), (01,01). Non-degenerate 2-simplices: (011,001) and (001,011). For comparison: in the pair (011,011), both components repeat at position 1, so (011,011) = s₁((01,01)) is degenerate; it lives on the diagonal.
Proposition 2.50 · Map and limits/colimits For a diagram F : I → sSet and every simplicial set X: Map(colim F(i), X) ≅ lim Map(F(i), X) and Map(X, lim F(i)) ≅ lim Map(X, F(i)). This is the key to using the decomposition Spn ≅ D¹ ⨿D⁰ ⋯ ⨿D⁰ D¹ in the Segal condition later on (Lemma 3.6).

sHoTT library · 02-simplicial-type-theory.rzk · RS17 §3Rzk counterparts: shapes as topes

In simplicial type theory (RS17), shapes are defined as topes (logical conditions) on cube variables: a point in the cube 2 is a variable t : 2, and a shape is a formula saying which points (t, s, ...) belong to it. The sHoTT-library counterparts of the sSet objects Dn, ∂Dn, L²₁:

Notes (sSet)Rzk (sHoTT)Tope condition
Δ¹\ t → TOP
Δ²\ (t , s) → s ≤ t
Δ³\ ((t1 , t2) , t3) → t3 ≤ t2 ∧ t2 ≤ t1
∂D¹∂Δ¹\ t → (t ≡ 0₂ ∨ t ≡ 1₂)
∂D²∂Δ²\ (t , s) → (s ≡ 0₂ ∨ t ≡ 1₂ ∨ s ≡ t)
L²₁ (= Sp²)Λ (and Λ²₁)\ (t , s) → (s ≡ 0₂ ∨ t ≡ 1₂)
L³₁, L³₂Λ³₁, Λ³₂inner horns; union of all faces except face k
D¹ × D¹Δ¹×Δ¹shape-prod 2 2 Δ¹ Δ¹
total boundary of the square∂□((∂Δ¹ t) ∧ (Δ¹ s)) ∨ ((Δ¹ t) ∧ (∂Δ¹ s))

Coordinate dictionary: the vertices of Δ² are 0 = (0₂,0₂), 1 = (1₂,0₂), 2 = (1₂,1₂). Edges: s ≡ 0₂ is edge 01, t ≡ 1₂ is edge 12, s ≡ t is the diagonal edge 02. Hence Λ = (edge 01) ∨ (edge 12), which is exactly the inner horn L²₁.

Tool · Tope explorer: which shapes contain the point (t, s)?

Click or drag a point in the square (points near the sides and the diagonal "snap" onto them). The tool shows whether the point satisfies each tope condition. Select a shape below to shade its region.

0 = (0₂,0₂) 1 = (1₂,0₂) 2 = (1₂,1₂) t → s →

The definitions in the library (02-simplicial-type-theory.rzk.md)

-- Simplices
#def Δ¹ : 2 → TOPE
  := \ t → TOP

#def Δ² : ( 2 × 2) → TOPE
  := \ (t , s) → s ≤ t

#def Δ³ : ( 2 × 2 × 2) → TOPE
  := \ ((t1 , t2) , t3) → t3 ≤ t2 ∧ t2 ≤ t1

-- Boundaries
#def ∂Δ¹ : Δ¹ → TOPE
  := \ t → (t ≡ 0₂ ∨ t ≡ 1₂)

#def ∂Δ² : Δ² → TOPE
  := \ (t , s) → (s ≡ 0₂ ∨ t ≡ 1₂ ∨ s ≡ t)

-- The 2-dimensional inner horn
#def Λ : ( 2 × 2) → TOPE
  := \ (t , s) → (s ≡ 0₂ ∨ t ≡ 1₂)

-- Products: the product of topes defines the product of shapes
#def shape-prod
  ( I J : CUBE)
  ( ψ : I → TOPE)
  ( χ : J → TOPE)
  : ( I × J) → TOPE
  := \ (t , s) → ψ t ∧ χ s

#def Δ¹×Δ¹ : ( 2 × 2) → TOPE
  := shape-prod 2 2 Δ¹ Δ¹

-- The total boundary of the square
#def ∂□ : ( 2 × 2) → TOPE
  := \ (t , s) → ((∂Δ¹ t) ∧ (Δ¹ s)) ∨ ((Δ¹ t) ∧ (∂Δ¹ s))

-- Vertical / horizontal boundary
#def ∂Δ¹×Δ¹ : ( 2 × 2) → TOPE
  := shape-prod 2 2 ∂Δ¹ Δ¹

#def Δ¹×∂Δ¹ : ( 2 × 2) → TOPE
  := shape-prod 2 2 Δ¹ ∂Δ¹

Face and degeneracy operations in Rzk / RS17 language

In RS17 §3, face and degeneracy operations are λ-terms on extension types: a face is obtained by fixing one coordinate on the boundary; a degeneracy by ignoring a variable. For example, the two degenerate 2-simplices associated to a 1-simplex:

-- degenerate 2-simplices from a 1-simplex (one variable is ignored)
λf. λ⟨t1, t2⟩. f(t1) : (Δ¹ → A) → (Δ² → A)
λf. λ⟨t1, t2⟩. f(t2) : (Δ¹ → A) → (Δ² → A)

-- the four 2-simplex faces of a 3-simplex (0 ≡ t3, t3 ≡ t2, t2 ≡ t1, t1 ≡ 1)
λf. λ⟨t1, t2⟩. f⟨t1, t2, 0⟩  : (Δ³ → A) → (Δ² → A)
λf. λ⟨t1, t2⟩. f⟨t1, t2, t2⟩ : (Δ³ → A) → (Δ² → A)
λf. λ⟨t1, t2⟩. f⟨t1, t1, t2⟩ : (Δ³ → A) → (Δ² → A)
λf. λ⟨t1, t2⟩. f⟨1, t1, t2⟩  : (Δ³ → A) → (Δ² → A)

Compare: on the sSet side, sᵢ repeats an element of the chain; on the Rzk side, the corresponding degeneracy ignores the relevant cube variable (it stays constant). On the sSet side, dⁱ skips an element; on the Rzk side, a face adds a coordinate constraint.

RS17 Prop. 3.6 · The square and its two triangles, on the Rzk side The sHoTT counterpart of the fact from Section 6 that "the square = two triangles glued along the diagonal": Δ¹ × Δ¹ behaves like the pushout of two copies of Δ² along their common diagonal boundary Δ¹₁ := {⟨t, s⟩ : 2 × 2 | t ≡ s}; in the library this is formalized as the statement that maps out of Δ² are a retract of maps out of Δ¹×Δ¹:
-- RS17, Proposition 3.6
#def Δ²-is-retract-Δ¹×Δ¹
  ( A : U)
  : is-retract-of (Δ² → A) (Δ¹×Δ¹ → A)
  :=
    ( ( \ f → \ (t , s) →
        recOR
          ( t ≤ s ↦ f (t , t)
          , s ≤ t ↦ f (t , s)))
    , ( ( \ f → \ ts → f ts) , \ _ → refl))

recOR splits the square into the regions t ≤ s and s ≤ t (the two triangles) and glues the definition piecewise; the overlap is the diagonal. Compare with the square figure in Section 6.

Exercise · Reading topes

On Δ², which edge does each of the three disjuncts in the condition of ∂Δ² correspond to? (Write your answers as chains.)

Show solution
The vertices are 0 = (0₂,0₂), 1 = (1₂,0₂), 2 = (1₂,1₂). The edge s ≡ 0₂ joins 0 and 1: 01. The edge t ≡ 1₂ joins 1 and 2: 12. The diagonal s ≡ t joins 0 and 2: 02. The union of the first two disjuncts is Λ = L²₁. Verify by selecting the shapes in the tope explorer.

Sources: N. Rasekh, ICERM School Notes: Complete Segal Spaces (v0.3), Section 2 · E. Riehl, M. Shulman, A type theory for synthetic ∞-categories (RS17), §3 · the sHoTT library, src/simplicial-hott/02-simplicial-type-theory.rzk.md. All definitions and numbering on this page are taken from these sources.

Section 3.1 · Notes: Def. 3.1 – Prop. 3.4The nerve construction

Definition 3.1 / Remark 3.2 · Nerve Let N : Cat → sSet be the functor given by the restricted Yoneda embedding applied to the functor ∆ → Cat. Explicitly:

• n = 0: NC₀ = Fun([0], C) = ObjC, the set of objects.

• n = 1: NC₁ = Fun([1], C) = MorC, the set of morphisms.

• n ≥ 2: NCn = Fun([n], C), the set of strings of n composable morphisms x₀ →f₁ x₁ →f₂ ⋯ →fn xn.

Visual reading of an n-simplex of NC: a string of n composable morphisms is exactly the data of the spine of an n-simplex whose vertices are the objects x₀, ..., xn. The rest of the simplex — the edges a₀a₁ with a₁ − a₀ ≥ 2 and the higher faces — is filled in by composites. For n = 2 this is a commuting triangle:

x₀ x₁ x₂ f₁ f₂ f₂∘f₁
A 2-simplex (f₁, f₂) ∈ NC₂: the spine (highlighted) is the data; the 02 edge is the composite (cf. Exercise 3.3).
0 1 2 0≤1 1≤2 0≤2
Lemma 3.3: N([n]) ≅ Dn. The nerve of the poset [2] is exactly the standard 2-simplex.
How to read δ: 02, 03, 001, ... For σ = (f₁, ..., fn) ∈ NCn and a chain δ = a₀...am : [m] → [n], the restriction δ(σ) ∈ NCm is the string whose j-th morphism is the composite of the segment from vertex aj−1 to vertex aj:
δ(σ) = (g₁, ..., gm),   gj = faj ∘ ⋯ ∘ faj−1+1 : xaj−1 → xaj
with the convention that an empty segment (aj−1 = aj) gives an identity. Special cases used constantly in the notes: 0(f) = source, 1(f) = target, 00(x) = idx, 02(f, g) = g ∘ f, 001(f) = (idx, f), 011(f) = (f, idy). These are exactly the identities used in the proof of Proposition 3.4 (fully faithfulness of N): F(idx) = f₁(00(x)) and F(h ∘ g) = 02(f₂(g, h)).
Tool · Nerve restriction calculator (symbolic)

Take σ = (f₁, ..., fn) ∈ NCn, a string x₀ → x₁ → ⋯ → xn. Enter n and a chain δ = a₀...am with values in [n]; the tool computes δ(σ) symbolically. Try 02, 00, 001, 013, 03.

Enter a chain and press the button.
Tool · Concrete nerve elements: the group Z/N (cf. Exercise 3.13)

A group G, seen as a one-object category, has NGn = Gn. Take G = Z/N with addition. Enter N, a tuple (g₁, ..., gn), and a chain; the tool computes δ concretely, where composing means adding mod N. For example 02(g, h) = g + h.

Enter a tuple and a chain.
Proposition 3.4 · The nerve is fully faithful N : HomCat(C, D) → HomsSet(NC, ND) is a bijection. Given a simplicial map f : NC → ND, the functor F is recovered by F(x) := f₀(x) on objects and F(g) := f₁(g) on morphisms; the simplicial identities 0, 1 (sources/targets), 00 (identities) and 02 (composites) force F to be a functor. So no information is lost by passing from C to NC.
Exercise 3.2 / 3.3

Let f : x → y be a 1-simplex of NC and τ = (f, g) ∈ NC₂ a pair x → y → z. Choose the right answers:

Show solution
s₀ repeats vertex 0, inserting idx in front: s₀(f) = (idx, f); s₁ repeats vertex 1: s₁(f) = (f, idy). For τ = (f, g): 01(τ) = f, 12(τ) = g, and 02(τ) = g ∘ f — the 2-simplex τ encodes the composition by carrying the composite on its long edge.
Exercise 3.1 / 3.13 · Faces of a 3-simplex

Let σ = (f₁, f₂, f₃) ∈ NC₃ (equivalently, (g, h, k) ∈ NG₃ for a group G). Match each face with its value:

Show solution
dᵢ deletes vertex i: d₀ = 123 = (f₂, f₃); d₁ = 023 = (f₂∘f₁, f₃); d₂ = 013 = (f₁, f₃∘f₂); d₃ = 012 = (f₁, f₂). In the group case: d₀ = (h, k), d₁ = (hg, k), d₂ = (g, kh), d₃ = (g, h). Verify with the calculator above.
Exercise 3.15

Let C be the category x → y → z with the composite g ∘ f. Compute:

Show solution
NC₀ = {x, y, z}: 3. NC₁ = {idx, idy, idz, f, g, g∘f}: 6. NC₂ = composable pairs: counting (into b) × (out of b) over b = x, y, z gives 1·3 + 2·2 + 3·1 = 10. A 2-simplex is degenerate exactly when one of its two morphisms is an identity (it is s₀(u) or s₁(u)); the only pair with both non-identities is (f, g): 1 non-degenerate 2-simplex.

Section 3.1 (cont.) · Notes: Def. 3.5 – Cor. 3.18The Segal condition and Theorem 3.15

Definition 3.5 · Segal condition A simplicial set X satisfies the Segal condition if the precomposition map in : HomsSet(Dn, X) → HomsSet(Spn, X) along the spine inclusion in : Spn → Dn is an isomorphism for all n ≥ 2.

Three equivalent readings (Lemma 3.6, Prop. 3.7, Prop. 3.8): by Lemma 3.6 and Yoneda, maps out of the spine are tuples, so the Segal condition says the map Xn → X₁ ×X₀ ⋯ ×X₀ X₁ (restricting an n-simplex to its spine edges 01, 12, ..., (n−1)n) is a bijection; equivalently, every map Spn → X extends uniquely along in to Dn → X. In words: an n-simplex is exactly a string of n composable edges, with all composites determined.

Tool · Unique spine filling, step by step (n = 2)

Step through what the Segal condition provides. The data is a map Sp² → X, i.e. two composable edges; the condition fills the whole triangle uniquely, and the 02 edge of the filler is the composite.

x y z f g ?

Proposition 3.9 / Remark 3.11: the infinitely many conditions can be packaged into a single one — i₂ : Map(D², X) → Map(Sp², X) being an isomorphism of simplicial sets (and Sp² ≅ L²₁, so this is a statement about the inner horn). Proposition 3.13: every nerve NC satisfies the Segal condition. Proposition 3.14: N(C × D) ≅ NC × ND.

Theorem 3.15 A simplicial set X is isomorphic to the nerve of a category if and only if X satisfies the Segal condition.

The proof of Theorem 3.15, visually

Given X Segal, write µ₂ : X₁ ×X₀ X₁ → X₂ and µ₃ : X₁ ×X₀ X₁ ×X₀ X₁ → X₃ for the inverse Segal maps, and build a category C: objects ObjC = X₀; morphisms HomC(x, y) = the fiber of (0, 1) : X₁ → X₀ × X₀ over (x, y); identity idx = 00(x); composition g ∘ f = 02µ₂(f, g). The category axioms are then read off from well-chosen simplices:

Tool · Associativity via the 3-simplex µ₃(f, g, h)

The tetrahedron below is µ₃(f, g, h) ∈ X₃: the unique 3-simplex with spine f, g, h. Click a face dᵢ to highlight it and see which equation it witnesses. All four faces belong to the same 3-simplex, so both bracketings compute the same 03 edge: (h∘g)∘f = 03µ₃(f, g, h) = h∘(g∘f).

x y z w f (01) g (12) h (23) g∘f (02) h∘g (13) 03
Click a face to see the equation it witnesses. The chain from the notes: (h∘g)∘f = 02*µ₂(f, 02*µ₂(g,h)) = 03*µ₃(f,g,h) = 02*µ₂(02*µ₂(f,g), h) = h∘(g∘f).

Unitality via degenerate triangles

For f : x → y, the degenerate 2-simplex 001(f) = s₀(f) has faces 01 = idx, 12 = f, 02 = f; reading off its 02 edge gives f ∘ idx = 02µ₂(00(x), f) = f. Similarly 011(f) = s₁(f) gives idy ∘ f = f. (In the notes' proof: f ∘ idx = 02µ₂(00(x), f) = f and idy ∘ f = 02µ₂(f, 00(y)) = f.)

x x y id_x f f s₀(f)
001(f) = s₀(f) witnesses f ∘ idx = f.
x y y f id_y f s₁(f)
011(f) = s₁(f) witnesses idy ∘ f = f.

Finally, NC ≅ X: NC₀ = X₀ and NC₁ = X₁ by construction, and for n ≥ 2 both NCn and Xn are identified with the same iterated pullback X₁ ×X₀ ⋯ ×X₀ X₁ — NC because nerves are Segal (Prop. 3.13), X by assumption. Remark 3.16: HomC(x, y) is recovered from NC as the fiber of (0, 1). Corollary 3.18: N : Cat → sSetSeg is an equivalence of categories — category theory is the study of Segal simplicial sets.

Rzk counterpart · 05-segal-types.rzk.md In sHoTT, the inner horn Λ (= L²₁) and the restriction map give the Segal condition as a locality condition, and unitality/associativity are proved from it (RS17 §5):
-- restriction of a 2-simplex to its underlying horn
#def horn-restriction (A : U)
  : ( Δ² → A) → (Λ → A)
  := \ f t → f t

-- Segal types as types local for Λ ⊂ Δ²
#def is-local-horn-inclusion : U → U
  := is-local-type (2 × 2) Δ² (\ t → Λ t)

-- unitality, RS17 Prop 5.8; in ∘ notation this says id_y ∘ f ≃ f
-- (comp-is-segal is written in diagrammatic order: comp ... f g means g ∘ f,
--  so the library calls this the "right unit law"; the mirror case
--  f ∘ id_x ≃ f is id-comp-is-segal)
#def comp-id-is-segal (A : U) (is-segal-A : is-segal A) (x y : A) (f : hom A x y)
  : ( comp-is-segal A is-segal-A x y y f (id-hom A y)) = f

-- associativity, RS17 Prop 5.9
#def associative-is-segal (A : U) (is-segal-A : is-segal A) (w x y z : A)
  ( f : hom A w x) (g : hom A x y) (h : hom A y z)
  : ( comp-is-segal A is-segal-A w y z (comp-is-segal A is-segal-A w x y f g) h)
  = ( comp-is-segal A is-segal-A w x z f (comp-is-segal A is-segal-A x y z g h))

Note the parallel: uniqueness of horn fillers replaces the bijection X₂ ≅ X₁ ×X₀ X₁, and the associativity proof again passes through Δ³ (RS17 Prop. 5.9 uses the "middle shuffle" Δ³ → Δ² × Δ¹).

Exercise · Which face witnesses which equation?

In the tetrahedron µ₃(f, g, h), match each face with what it witnesses:

Show solution
d₃ = 012 contains edges 01 = f, 12 = g, 02 — it witnesses g∘f. d₀ = 123 witnesses h∘g on the 13 edge. d₂ = 013 contains 01 = f, 13 = h∘g, 03 — it witnesses (h∘g)∘f. d₁ = 023 contains 02 = g∘f, 23 = h, 03 — it witnesses h∘(g∘f). Same 03 edge, hence associativity.

Section 3.2 · Notes: Def. 3.19 – Prop. 3.36From algebraic topology: homotopy, CW-complexes, the topological simplex, Sing

Definition 3.19 / 3.21 / 3.24 · Homotopy and weak homotopy equivalence Two continuous maps f, g : X → Y are homotopic if there is a continuous H : X × [0, 1] → Y with H|X×{0} = f and H|X×{1} = g. For a pointed space (X, x), the n-th homotopy set πn(X, x) is the set of homotopy classes of pointed maps Sn → X. A map f is a weak homotopy equivalence if it induces bijections on π₀ and on all πn at all basepoints.
X × {0} : f X × {1} : g X × [0,1] H Y
A homotopy is a map out of the cylinder X × [0, 1], restricting to f and g on the two ends.
Definition 3.25 – Theorem 3.29 · CW-complexes and Whitehead A CW-complex is built inductively: X⁰ is a discrete set of points, and Xn is obtained from Xn−1 by attaching n-cells along maps ∂Dn → Xn−1. The CW-approximation theorem (3.26) replaces any space by a weakly equivalent CW-complex, and Whitehead's theorem (3.29) says that between CW-complexes, weak homotopy equivalences are already homotopy equivalences (Lemma 3.28 gives the easy converse). So up to weak equivalence, homotopy theory happens on CW-complexes.
X⁰: one 0-cell attach one 1-cell: S¹ attach a 2-cell: D²
Cell-by-cell construction: the inductive structure is what makes CW-complexes computable.
Definition 3.30 / Lemma 3.31 · The topological n-simplex |Dn| := {(t₀, ..., tn) ∈ Rn+1 | Σ tᵢ = 1, tᵢ ≥ 0}. The assignment [n] ↦ |Dn| is a functor |D| : ∆ → Top, with
di(t₀, ..., tn) = (t₀, ..., ti−1, 0, ti, ..., tn)   ·   si(t₀, ..., tn) = (t₀, ..., ti−1, ti + ti+1, ti+2, ..., tn)
So di inserts a zero (the face where the i-th coordinate vanishes) and si merges two coordinates (collapsing onto a lower simplex) — the continuous shadows of "skip an element" and "repeat an element".
Tool · Barycentric coordinates on |D²|

Click or drag inside the triangle. The point is displayed in barycentric coordinates (t₀, t₁, t₂) with t₀ + t₁ + t₂ = 1. The face tᵢ = 0 is the edge opposite vertex i — exactly the image of di.

e₀ e₁ e₂ t₂ = 0 t₀ = 0 t₁ = 0
Tool · dⁱ and sⁱ on coordinate tuples (Lemma 3.31)

Enter a point of |Dn| as a comma-separated tuple summing to 1, choose i, and apply di (insert 0) or si (merge tᵢ + ti+1).

Enter a tuple and press a button.
Definition 3.32 – Proposition 3.36 · Sing and simplicial homotopy Sing : Top → sSet is the restricted Yoneda embedding applied to |D|: explicitly Sing(X)n = HomTop(|Dn|, X) — vertices are points, edges are paths, 2-simplices are maps of a solid triangle into X. Two simplicial maps f, g : X → Y are homotopic (Def. 3.34) if there is H : X × D¹ → Y restricting to f and g on the two ends — the same cylinder picture, with D¹ replacing [0, 1]; homotopy equivalence (Def. 3.35) is defined as usual. Proposition 3.36: Sing sends homotopic maps to homotopic maps and preserves homotopy equivalences.
Exercise 3.18

|D⁰| is the point (1) ∈ R¹, |D¹| the segment in R², |D²| the triangle in R³. Compute, using the formulas of Lemma 3.31:

Show solution
d⁰ inserts a 0 at position 0: d⁰(1) = (0, 1); d¹ inserts it at position 1: d¹(1) = (1, 0). s⁰ merges: s⁰(t₀, t₁) = (t₀ + t₁) = (1). You can verify with the coordinate tool above.
Exercise 3.19

Let S be a discrete topological space. What is Sing(S)n, and what kind of simplicial set is Sing(S)?

Show solution
|Dn| is connected, so a continuous map |Dn| → S into a discrete space is constant; hence Sing(S)n ≅ S for all n, with all face and degeneracy maps identities — the constant simplicial set associated to S (cf. Definition 2.14).

Section 3.2 (cont.) · Notes: Def. 3.37 – Rem. 3.46Kan complexes, π₀, and the homotopy theorem

Definition 3.37 · Kan complex A simplicial set K is a Kan complex if for every n ≥ 1 and 0 ≤ k ≤ n, every map Lnk → K extends to a map Dn → K; equivalently, Hom(Dn, K) → Hom(Lnk, K) is surjective. Note the contrast with the Segal condition: all horns (including the outer ones Ln₀, Lnn), and only existence of fillers, not uniqueness.
Lnk K Dn
The Kan lifting condition: every horn in K fills.

Proposition 3.38: For every topological space X, Sing(X) is a Kan complex — |Lnk| is a retract of |Dn|, so continuous horns always fill. This gives a functor Sing : Top → Kan (Def. 3.39), which is not an equivalence of categories, but becomes one after passing to homotopy.

Definition 3.40 · Path-components For a simplicial set S, define x ∼ y on S₀ if there is an edge f ∈ S₁ with 0(f) = x and 1(f) = y. Then π₀(S) = S₀ / ∼ is the quotient by the equivalence relation generated by ∼.
Lemma 3.41 · For a Kan complex, ∼ is already an equivalence relation The proof is pure horn filling, and it is worth seeing: symmetry and transitivity are produced by filling an outer and an inner 2-horn.
Tool · The proof of Lemma 3.41, by horn filling

Here the triangle lives in a Kan complex K; the vertices are vertices of K and the edges are edges (1-simplices) between them. Pick a case and step through the horn, the filler, and the output face.

x y x f ? id

Lemma 3.42: π₀(Sing(X)) ≅ π₀(X) — vertices of Sing(X) are points, edges are paths, so the simplicial π₀ recovers path-components. Combined with Def. 3.43 (Ho(TopCW)) and Def. 3.44 (Ho(Kan)):

Theorem 3.45 · [GJ09, Theorem 11.4] The functor Sing : Ho(TopCW) → Ho(Kan) is an equivalence of categories. Together with Corollary 3.18, Section 3 has thus embedded both classical worlds into sSet: categories as Segal simplicial sets, and (the homotopy theory of) CW-complexes as Kan complexes.

Remark 3.46: one consequence is the 2-out-of-3 property for homotopy equivalences of Kan complexes: if two of f, g, g ∘ f are homotopy equivalences, so is the third.

Outlook · Lemma 4.1: the two worlds barely intersect NC is a Kan complex if and only if C is a groupoid: outer horn filling is invertibility — filling L²₀ asks for a left inverse, filling L²₂ for a right inverse (spelled out in lifting language in section 14). This near-empty intersection between sSetSeg and Kan is what motivates simplicial spaces (sections 12–13).
Exercise · Reading the proof of Lemma 3.41

Match each step of the proof with the horn and output it uses (H : f ⇒ g a homotopy, i.e. an edge in a mapping object):

Show solution
Reflexivity: the constant map D¹ → K. Symmetry: σ : L²₀ → K with 01 ↦ H and 02 ↦ idf; a Kan filler H′ : D² → K exists, and 12H′ is the homotopy g ⇒ f. Transitivity: H₁ + H₂ : L²₁ → K with 01 ↦ H₁, 12 ↦ H₂; fill to H₁₂ and restrict along 02. Step through the tool above to see both.
Exercise · π₀ practice (cf. Exercise 5.23)

Compute the number of path-components:

Show solution
D¹ has the edge 01 connecting its two vertices: π₀(D¹) has 1 element. ∂D¹ has vertices 0, 1 and no edge between them: 2 elements. ∂D² has three vertices all connected by edges: 1 element. D⁰ ⨿ D⁰: 2 elements.

Section 4 · Notes: Def. 4.2 – Lemma 4.14Two directions: simplicial spaces and the bisimplicial grid

Definition 4.2 / Remark 4.3 · Simplicial space A simplicial space is a functor X : ∆op → sSet; the category is sS := Fun(∆op, sSet) ≅ Fun(∆op × ∆op, Set). So X is a grid of sets Xnk with two directions: the first index n runs horizontally (this will be the categorical direction), the second index k runs vertically (the spatial direction). Each column Xn• is a simplicial set.
Definitions 4.4 / 4.6 · The two inclusions of sSet

disc : sSet → sS (categorical / horizontal inclusion): (disc S)nk = Sn — all vertical maps are identities (Intuition 4.5).

const : sSet → sS (spatial / vertical inclusion): (const S)nk = Sk — all horizontal maps are identities (Intuition 4.7).

Same simplicial set, two orthogonal placements in the grid. Exercise 4.9: if S is a constant simplicial set, disc S ≅ const S.

Tool · The bisimplicial grid: where does the data live?

Pick a simplicial space; each cell shows the cardinality |Xnk|, shaded by size. For disc the columns are constant (the data varies horizontally), for const the rows are constant (the data varies vertically) — you see the two embeddings as two stripe patterns. Click a cell for its identification and the probe that reads it (Lemma 4.11).

n → categorical (horizontal) · k ↓ spatial (vertical)
Notation 4.10 / Warning 0.1 · The notation switchboard The representables of sSet acquire two names in sS, depending on the direction they are placed in:
sSetcategorical direction (disc)spatial direction (const)
Dnn := disc(Dn)Dl := const(Dl)
∂Dn∂∆n := disc(∂Dn)∂Dl := const(∂Dl)
LnkΛnk := disc(Lnk)
SpnSpn := disc(Spn) (n ≥ 2)

Rule of thumb: ∆-letters live in the categorical direction, D-letters in the spatial one. A general simplicial set S is silently identified with const S. This is the second bullet of Warning 0.1, again chosen to match Rzk.

Lemma 4.11 / Lemma 4.14 · Bisimplicial Yoneda HomsS(∆n × Dl, X) ≅ Xnl: the product ∆n × Dl is the probe that reads the single cell (n, l) of the grid. In particular HomsS(∆n, X) ≅ Xn0 (Exercise 4.11), and at the level of simplicial sets, Map(∆n, X) ≅ Xn reads a whole column (Lemma 4.14).
Exercise 4.2

Compute the four sets (disc D¹)nk and compare them with (const D¹)nk. Enter the cardinalities:

disc D¹:
const D¹:
Show solution
(disc D¹)nk = D¹n: the values 2, 3, 2, 3 depend only on n. (const D¹)nk = D¹k: the values 2, 2, 3, 3 depend only on k. (|D¹0| = |{0, 1}| = 2, |D¹1| = |{00, 01, 11}| = 3.) Select disc D¹ and const D¹ in the grid tool and compare the stripe patterns.
Exercise 4.3 / 4.11

Identify the level sets of the representables:

Show solution
(∆ⁿ)kl = (disc Dⁿ)kl = Dⁿk = Hom([k], [n]) — independent of l. (Dl)kn = (const Dl)kn = Dln = Hom([n], [l]) — independent of k. And HomsS(∆ⁿ, X) ≅ Xn0: since ∆ⁿ = ∆ⁿ × D⁰, this is the case l = 0 of Lemma 4.11.

Section 6 · Notes: Def. 6.1 – Ex. 6.11Reedy fibrancy: making the vertical direction spatial

The spatial direction of a simplicial space should behave like spaces, i.e. Kan complexes. But a naive level-wise Kan condition is not enough: the Segal condition will be a statement about pullbacks like X₁ ×X₀ X₁, and, as Section 5 shows (Lemma 5.12), such pullbacks behave homotopy-invariantly when the maps 0, 1 : X₁ → X₀ are Kan fibrations. Reedy fibrancy packages exactly this dependency.

Definition 6.1 / Lemma 6.2 · Reedy fibrancy A map of simplicial spaces p : Y → X is a Reedy fibration if for every n ≥ 0, k ≥ 1, 0 ≤ i ≤ k the pushout-product inclusion ∂∆n × Dk ⨿∂∆n×Lkin × Lki ↪ ∆n × Dk lifts against p; X is Reedy fibrant if X → ∆⁰ is a Reedy fibration. The usable characterization (Lemma 6.2): X is Reedy fibrant iff for every n ≥ 0 the map Map(∆n, X) ↠ Map(∂∆n, X) is a Kan fibration.

What Reedy fibrancy buys, on the grid (Examples 6.3 – 6.5): every column Xn is a Kan complex; the map (0, 1) : X₁ → X₀ × X₀ is a Kan fibration (so its fibers — the future mapping spaces — are Kan complexes, and pullbacks over X₀ are homotopy invariant); more generally (0, ..., n) : Xn → (X₀)n+1 is a Kan fibration.

X₀ X₁ X₂ Kan cplx Kan cplx Kan cplx (0*,1*) Kan fib. Ex. 6.3: columns Kan Ex. 6.4/6.5: vertex maps are fibrations
A Reedy fibrant simplicial space: Kan columns, dependently (as fibrations).

disc always works, const usually fails. Proposition 6.8: disc(A) is Reedy fibrant for every simplicial set A — the mapping spaces Map(∆n, disc A) and Map(∂∆n, disc A) are constant simplicial sets (Lemmas 6.6, 6.7), and any map between constant simplicial sets is a Kan fibration (Lemma 5.26). In particular ∆n, ∂∆n, Spn are Reedy fibrant (Example 6.9). On the const side, even a Kan complex can fail:

Examples 6.10 / 6.11 · The two counterexamples

• const(D¹) is not Reedy fibrant: its column 0 is D¹ = N([1]), which would have to be a Kan complex — but by Lemma 4.1 that would force [1] to be a groupoid.

• const(N(I(1))) is not Reedy fibrant, even though N(I(1)) is a Kan complex (Lemma 4.1). For const S the map Map(∆¹, const S) → Map(∂∆¹, const S) is the diagonal ∆ : S → S × S (Exercise 6.16), and by Example 6.4 it would have to be a Kan fibration. The obstruction is explicit (Exercise 6.17): in the lifting problem below, a lift would be an edge e of N(I(1)) with (e, e) = (01, 00) — an edge equal to 01 and 00 at once.

L¹₀ ≅ D⁰ N(I(1)) N(I(1)) × N(I(1)) 0 (01, 00)

Contrast with disc(D¹): its columns are the constant simplicial sets on D¹₀, D¹₁, ..., all Kan — even though D¹ itself is not a Kan complex (Exercise 6.19). Reedy fibrancy tests the two embeddings very differently (Exercise 6.4).

Exercise · Which simplicial spaces are Reedy fibrant?

Decide for each:

Show solution
∆², Sp₃, disc(N(I(1))): Reedy fibrant — disc(A) always is (Prop. 6.8, Ex. 6.9). const(D⁰): Reedy fibrant — D⁰ is a constant simplicial set, so const(D⁰) ≅ disc(D⁰) (Ex. 4.9), and this is ∆⁰. const(D¹): not — column 0 is N([1]), not Kan (Ex. 6.10). const(N(I(1))): not — the diagonal fails to be a Kan fibration (Ex. 6.11), despite N(I(1)) being Kan. The moral: on the const side, Reedy fibrancy sees homotopical structure that "being a Kan complex level-wise" misses.

Section 5 · Notes: Def. 5.6 – Ex. 5.28Kan fibrations: dependent spaces

Why fibrations? Pullbacks of Kan complexes are not homotopy invariant. Example 5.5 is the smallest possible failure: replace N(I(1)) by the equivalent D⁰ and the pullback jumps from ∅ to a point.

D⁰ N(I(1)) D⁰ 0 1 D⁰ D⁰ D⁰ = = pullback: ∅ pullback: D⁰ not ≃
Example 5.5: all vertical maps are homotopy equivalences (0 : D⁰ → N(I(1)) is one by Example 5.3), yet the induced map of pullbacks is ∅ → D⁰.
Definition 5.6 / Remark 5.8 · Kan fibration A map of simplicial sets Y → X is a Kan fibration if for every n ≥ 1, 0 ≤ i ≤ n, every square Lnᵢ → Y, Dn → X admits a lift Dn → Y (in diagrams: ↠). This is the relative form of the Kan condition: a Kan fibration Y ↠ D⁰ is precisely a Kan complex (Remark 5.8). Fibrations compose and are stable under pullback (Lemmas 5.9, 5.10), and — this is the payoff — if 0, 1 : X₁ → X₀ are Kan fibrations, then pullbacks like X₁ ×X₀ X₁ behave homotopy-invariantly (Lemma 5.12, as used in Section 6).
Lⁿᵢ Y Dⁿ X incl. Kan fib. ∃ lift
The lifting square: the horn is already mapped into Y, the whole simplex only into X; a Kan fibration promotes it to a map into Y, over X. Taking X = D⁰ recovers the Kan condition.
Lemma 4.1 in lifting language · Two remarks

Outer horn filling in a nerve is the search for inverses: filling L²₀ (edges 01 = f, 02 = id) asks for a left inverse of f, filling L²₂ for a right inverse. So neither fills in N[1] (nothing maps back in [1]), while both fill in N(I(1)) — this is the proof of Lemma 4.1.

Being a fibration is a property of the map, not of the objects (Exercise 5.25): N(0) : D⁰ → N(I(1)) has Kan source and Kan target, yet is not a Kan fibration — any lift D¹ → D⁰ of the edge 01 would be constant.

Definition 5.18 / Proposition 5.15 · Path spaces For a Kan complex K and vertices x, y, the path space PathK(x, y) is the fiber of Map(D¹, K) → K × K over (x, y); by Proposition 5.15 this map is a Kan fibration, so path spaces are the first examples of well-behaved fibers. Its vertices are exactly the edges of K from x to y.

Section 7 · Notes: Def. 7.1 – Rem. 7.16Uniqueness vs. contractibility

Definitions 7.1 / 7.3 / 7.6 · The dictionary In set theory, a property φ on X carves out a subset {x ∈ X | φ(x)}, and x₀ satisfies φ uniquely if that subset is a singleton (Definition 7.1) — e.g. the composite g ∘ f is the unique composition of f and g (Example 7.2). The homotopical translation replaces each word: sets ↝ Kan complexes, subsets ↝ sub-Kan complexes L ⊆ K (Definition 7.5), singleton ↝ contractible — K is contractible if the unique map K → D⁰ is a homotopy equivalence (Definition 7.3). A homotopical property L ⊆ K is homotopically unique if L is contractible (Definition 7.6).

Local vs. global, and why fibrancy is needed. For sets, a map is a bijection iff every pre-image is a singleton (Lemma 7.7): global uniqueness = local uniqueness at every point (Intuition 7.8). The naive homotopical analogue is false — Example 7.9: 0 : D⁰ → N(I(1)) is a homotopy equivalence, but the pre-image of the vertex 1 is ∅, which is not contractible (Exercise 7.9). Global equivalence does not see the fibers unless the map is a fibration:

Definition 7.10 / Proposition 7.12 · Trivial fibrations A trivial fibration is a Kan fibration that is also a homotopy equivalence. The three faces of the concept (Proposition 7.12) — for f : Y → X the following are equivalent:

(1) f is a trivial fibration;

(2) f is a Kan fibration and every fiber f−1(x) is a contractible Kan complex — dependent homotopical uniqueness;

(3) every square ∂Dn → Y, Dn → X (n ≥ 0) admits a lift — boundary filling.

Remark 7.11: over the point, a trivial fibration K ↠ D⁰ is exactly a contractible Kan complex — the definition simultaneously generalizes contractibility and dependence. Trivial fibrations are stable under pullback and composition (Lemmas 7.13, 7.14).

Horn filling vs. boundary filling

The two lifting conditions of this module, side by side, at n = 2: a Kan fibration fills horns (one boundary face missing — existence of composites, inverses, ...); a trivial fibration fills boundaries (only the interior missing — any already-given candidate boundary is realized, which is where uniqueness-up-to-homotopy comes from).

Kan fibration: horn filling. L²₁ ↪ D²: the edge 02 and the interior are produced.
Trivial fibration: boundary filling — a strictly stronger condition. ∂D² ↪ D²: only the interior is produced.
Exercise 7.5 / 7.12 / 7.13 · Reading the low cases of boundary filling

In Proposition 7.12(3), unpack the cases n = 0 and n = 1, using ∂D⁰ = ∅ and ∂D¹ ≅ D⁰ ⨿ D⁰:

Show solution
n = 0: the boundary ∂D⁰ is empty, so the square is just a vertex of X, and the lift is a pre-image vertex: f is surjective on 0-simplices (Exercise 7.12). n = 1: the boundary data is a pair of vertices of Y over the endpoints of an edge of X, and the lift is an edge of Y joining them over it. In particular, taking f(y₀) = f(y₁) = x over the degenerate edge, one gets a 1-simplex in the fiber f−1(x) connecting y₀ and y₁ (Exercise 7.13) — fibers are connected in the way contractibility demands.
Exercise 7.21 / 7.22 · Contractibility in the simplest cases

Let S be a set, seen as a constant simplicial set. When is S contractible?

Now take S = {a, b}. Its sub-Kan complexes are ∅, {a}, {b}, {a, b}. Tick the contractible ones:


Show solution
A constant simplicial set S is contractible exactly when S has one element — this is Remark 7.4 made literal: contractible Kan complexes are the homotopical singletons. For S = {a, b}: {a} and {b} are contractible; {a, b} maps to D⁰ non-injectively on components; and ∅ is not contractible (Exercise 7.9 — the map ∅ → D⁰ has no section).
Remark 7.16 · The bridge to HoTT Categories are usually defined set-theoretically, where uniqueness means singletons — a good match for classical category theory. This section shows that homotopical uniqueness is more intricate in that foundation: it requires contractibility of Kan complexes and, dependently, trivial fibrations. Remark 7.16 asks whether the foundation itself can be adjusted so that its native notion of uniqueness is homotopical uniqueness — and notes this has been achieved by homotopy type theory. This is exactly how sHoTT/Rzk will phrase everything ahead: in RS17, "the composite is unique" is literally the statement that a type (of fillers) is contractible.

Section 8 · Notes: Def. 8.2 – Lemma 8.26Segal spaces: composition as a contractible choice

Reedy fibrancy (section 13) makes a simplicial space behave like a space vertically; now we make it behave like a category horizontally. The key property was the Segal condition (Theorem 3.15) — here is its homotopical analogue.

Lemma 8.1 / Definition 8.2 · Segal space For any simplicial space X there is a natural isomorphism Map(Spn, X) ≅ X₁ ×X₀ ⋯ ×X₀ X₁ (n factors, pullbacks over 0, 1). A Segal space is a Reedy fibrant simplicial space T such that for every n ≥ 2 the Kan fibration
in : Tn ≅ Map(∆n, T) ↠ Map(Spn, T) ≅ T₁ ×T₀ ⋯ ×T₀ T₁
is a trivial Kan fibration. Reedy fibrancy is what makes the right-hand side homotopy invariant (Remark 8.3, via Lemma 5.12), and by Proposition 8.5 the single map i₂ : ⟦∆², T⟧ ↠ ⟦Sp₂, T⟧ being a trivial Reedy fibration packages all n at once. In the language of section 15: the space of fillers is homotopically unique (Intuition 8.6).
Figure · Strict vs. homotopical: one filler vs. a contractible space of fillers

Left: the world of Theorem 3.15 — the Segal condition on a simplicial set makes the fiber of i₂ over (f, g) a single point. Right: in a Segal space, the fiber is a contractible Kan complex — several fillers σ⁽¹⁾, σ⁽²⁾, σ⁽³⁾, whose composites are joined by homotopies in mapT(x, z) (Lemma 8.26, the strip below).

x y z f g g∘f σ (unique)
simplicial set + Segal condition:
the fiber over (f, g) is one point
x y z f g (g∘f)⁽¹⁾, (g∘f)⁽²⁾, (g∘f)⁽³⁾ σ⁽¹⁾, σ⁽²⁾, σ⁽³⁾ map_T(x, z)
Segal space: a contractible space of fillers;
choices of g∘f are homotopic in map_T(x, z)
Definitions 8.8 / 8.10 / 8.11 / 8.17 · The category theory begins Objects of a Segal space T: ObjT = T₀₀ (vertices of T₀). Morphisms: points of T₁; source 0f, target 1f. The mapping space mapT(x, y) is the fiber of (0, 1) : T₁ → T₀ × T₀ over (x, y) — a Kan complex, since that map is a Kan fibration (Lemma 8.12; compare Remark 3.16, where the same pullback recovered HomC). Two morphisms f, g are homotopic (f ≃ g) if there is a map D¹ → mapT(x, y) from f to g; since mapT(x, y) is Kan, this is an equivalence relation (Lemma 8.19) — the proof is exactly the horn-filling argument of Lemma 3.41, which you can step through in section 11. Generalized mapping spaces mapT(x₀, ..., xn) are defined analogously, and the Segal condition makes mapT(x₀, ..., xn) ↠ mapT(x₀, x₁) × ⋯ × mapT(xn−1, xn) a trivial Kan fibration (Proposition 8.22).
Definition 8.23 / Proposition 8.24 / Lemma 8.26 · The space of compositions For f : x → y, g : y → z, the space of compositions CompT(f, g) is the pullback of mapT(x, y, z) → mapT(x, y) × mapT(y, z) over the point (f, g). A vertex is exactly a 2-cell σ with 01σ = f and 12σ = g (Intuition 8.25) — a filled triangle witnessing "02σ is a composite of f and g". Proposition 8.24: CompT(f, g) is a contractible Kan complex — composition exists and is homotopically unique, but not literally unique. Lemma 8.26: any two choices of composition are homotopic in mapT(x, z), which is what the strip in the figure displays.

Section 8 (cont.) · Notes: Def. 8.31 – Prop. 8.45Identities, the homotopy category, and the Rzk dictionary

Definition 8.31 / Proposition 8.33 · Identities and the category laws, up to homotopy The identity of an object x is idx := 00(x) in T₁ — the same degenerate-edge recipe as in nerves; it has source and target x (Lemma 8.32). Proposition 8.33: composition is homotopically associative and unital: h ∘ (g ∘ f) ≃ (h ∘ g) ∘ f, and idy ∘ f ≃ f ≃ f ∘ idx.

The associativity picture is the homotopical upgrade of the tetrahedron from section 9: lift the triple (f, g, h) along the equivalence to a 3-cell σ in mapT(x, y, z, w); restricting along 013 computes (h∘g)∘f, restricting along 023 computes h∘(g∘f), and both land on the same 03-edge of the same σ.

σ ∈ map(x,y,z,w) x y z w f (01) g (12) h (23) g∘f (02) h∘g (13) 03*σ face 013 face 023
Prop. 8.33 (associativity) on the 3-cell σ: the teal face 013 has edges f, h∘g and computes (h∘g)∘f; the amber face 023 has edges g∘f, h and computes h∘(g∘f). Both are faces of the same σ, so both composites are the same long edge 03σ.
x x y id_x (01) f (12) f (02) 001*(f)=s₀(f)
Prop. 8.33 (unitality): the 2-cell 001(f) has faces idx, f, f (Ex. 8.17), so f ∘ idx ≃ f; the mirror 011(f) handles idy ∘ f (Ex. 8.18).
Construction 8.42 / Theorems 8.43 – 8.45 · The homotopy category Ho(T): objects are the objects of T; HomHo(T)(x, y) := mapT(x, y)/≃; composition descends from the composition maps because homotopy respects composition (Lemmas 8.26, 8.27). This is a genuine category (Theorem 8.43) — well-definedness, associativity, and unitality are exactly Lemmas 8.26/8.27 and Proposition 8.33. Every functor of Segal spaces (a map of simplicial spaces) induces Ho(F) : Ho(T) → Ho(U) (Theorem 8.44), and fully faithful functors stay fully faithful (Proposition 8.45). So a Segal space remembers a category — plus all the homotopies the category forgets.
Rzk counterpart · Segal types are exactly this section RS17's definition reads: a Segal type is a type A in which every inner horn has a unique filler, meaning every extension type ⟨Δ² → A |Λ h⟩ is contractible — uniqueness stated as contractibility, exactly Definition 8.2 + Proposition 8.24 in the foundation promised by Remark 7.16. The library results line up one-to-one with this section (05-segal-types):
-- identity = degenerate arrow             (Def 8.31: id_x = 00*(x))
id-hom

-- unit law, RS17 Prop 5.8                 (Prop 8.33: id_y ∘ f ≃ f)
--   comp-is-segal is diagrammatic: comp ... f g = g ∘ f
--   the mirror case f ∘ id_x ≃ f is id-comp-is-segal
comp-id-is-segal

-- associativity, RS17 Prop 5.9            (Prop 8.33: (h∘g)∘f ≃ h∘(g∘f))
associative-is-segal

Section 9 · Notes: Def. 9.2 – Lemma 9.10Homotopy equivalences in Segal spaces

Definition 9.2 / Remark 9.3 · Homotopy equivalence A morphism f in mapT(x, y) is a homotopy equivalence if there are g, h in mapT(y, x) such that f ∘ g ≃ idy and h ∘ f ≃ idx. Equivalently (Remark 9.3): idy is a choice of composition of g and f, and idx a choice of composition of f and h.

The hidden data (Intuition 9.4): the definition looks like "three maps exist", but in a Segal space each homotopy to an identity is carried by a witnessing 2-cell. What the definition really provides is two filled triangles:

x y x f h id_x σ_h
σh: a 2-cell witnessing h ∘ f ≃ idx — h is a left inverse.
y x y g f id_y σ_g
σg: a 2-cell witnessing f ∘ g ≃ idy — g is a right inverse.

Exercise 9.1 records where these witnesses live: the right-inverse triangle is a point of a pullback of mapT(y, x, y), the left-inverse triangle of mapT(x, y, x).

Lemma 9.5: any left inverse h and any right inverse g of f agree up to homotopy — the classical chain g ≃ idx ∘ g ≃ (h ∘ f) ∘ g ≃ h ∘ (f ∘ g) ≃ h ∘ idy ≃ h, run entirely with Prop. 8.33 (in sHoTT this is RS17 Prop. 10.1).

Propositions 9.6 / 9.8 · Two characterizations

Via the homotopy category (Prop. 9.6): f is a homotopy equivalence iff [f] is an isomorphism in Ho(T). "Being an equivalence" carries no homotopical information beyond Ho(T) (Intuition 9.7).

Via mapping spaces (Prop. 9.8): f is a homotopy equivalence iff post-composition f : mapT(z, x) → mapT(z, y) is a homotopy equivalence for every z, iff pre-composition f : mapT(y, z) → mapT(x, z) is.

Definition 9.9 / Lemma 9.10: T is a Segal space groupoid if every morphism is a homotopy equivalence — equivalently, Ho(T) is a groupoid.

Section 9 (cont.) · Notes: Def. 9.31 – Lemma 9.46Property vs. structure: the space of equivalences

The three players

Thoequiv ⊆ T₁: the sub-simplicial set generated by the vertices that are homotopy equivalences (Exercise 9.2); it is an inclusion of path-components (Lemma 9.25). A point of it is a morphism f together with the property of being an equivalence.

Thoeqchoice: a point is a tuple (f, g, h, σg, σh) — the morphism plus chosen inverses plus the witnessing 2-cells (as in the proof of Theorem 9.43). This is structure.

LInv(f), RInv(f) (Definition 9.31): the fibers of (02, 01), (02, 12) : T₂ → T₁ ×T₀ T₁ over (idx, f) and (idy, f). A point of LInv(f) is a morphism h with a 2-cell from h∘f to idx; similarly for RInv(f) (Intuition 9.32) — exactly the two triangles of section 18.

Theorem 9.43: the forgetful map Pforgchoice : Thoeqchoice → Thoequiv is a homotopy equivalence — forgetting the chosen inverses loses nothing; the choice is homotopically unique.

Remark 9.42 · The hypothesis is doing real work Proposition 9.40 (LInv(f) and RInv(f) are contractible) assumes f is already a homotopy equivalence — and that is essential. Even in the category of sets, an injection has many left inverses and a surjection many right inverses: the space of one-sided inverses of a non-equivalence is not contractible. Complement (Exercise 9.33): if LInv(f) and RInv(f) are both merely non-empty, then f is a homotopy equivalence — so the whole picture switches on at once.
Definition 9.45 / Lemma 9.46 · The local version hoequivT(x, y) := the fiber of Thoequiv → T₀ × T₀ over (x, y) — isomorphic to the sub-simplicial set of mapT(x, y) generated by the homotopy equivalences. This is the object the completeness condition of Section 10 will compare against PathT₀(x, y).
Rzk counterpart · 10-rezk-types: "being an isomorphism is a proposition" RS17 §10.1 defines, for f : homA(x, y) in a Segal type,
isiso(f) := ( Σg : hom(y,x) g ∘ f = idx ) × ( Σh : hom(y,x) f ∘ h = idy )
— literally a point of Thoeqchoice. The library then proves this section line by line:
-- one two-sided inverse iff separate left and right inverses  (RS17 Prop 10.1 = Lemma 9.5)
#def inverse-iff-iso-arrow
  : iff (has-inverse-arrow A is-segal-A x y f) (is-iso-arrow A is-segal-A x y f)

-- being an isomorphism is a proposition  (RS17 Prop 10.2 = Thm 9.43 + Prop 9.40)
#def is-prop-is-iso-arrow
  : is-prop (is-iso-arrow A is-segal-A x y f)

The proof of is-prop-is-iso-arrow shows the retraction type and the section type are contractible — the exact counterparts of LInv(f), RInv(f) in Proposition 9.40, obtained as fibers of the pre/post-composition equivalences of Proposition 9.8. A property in HoTT is a type with contractible inhabitedness: Intuition 9.44, stated inside the theory.

Section 10 · Notes: Ex. 10.1 – Ex. 10.14Why Segal spaces are not enough: E(1)

Examples 10.1 / 10.2 · The two embeddings into Segal spaces

Categories: disc(NC) is a Segal space; its mapping spaces are constant on HomC(x, y) and Ho(disc NC) ≅ C (section 17).

Spaces: for a Kan complex K, const(K) is not Reedy fibrant (Example 6.11), but the equivalent ιK := Map(D, K) is a Segal space. Its objects are the points of K, its morphisms are paths D¹ → K, and mapιK(x, y) = PathK(x, y) (section 14). Composition is concatenation of paths — well-defined only up to homotopy, which is exactly why the contractibility of CompT(f, g) (Prop. 8.24) is the right axiom. Every morphism is a homotopy equivalence (the inverse path is Lemma 3.41's symmetry), so ιK is a Segal space groupoid, and Ho(ιK) = Π₁(K), the fundamental groupoid.

The problem, in one example. Let E(1) := disc(N(I(1))) (Notation 10.3). The functor 0 : [0] → I(1) is an equivalence of categories — yet ∆⁰ → E(1) is not an equivalence of Segal spaces: by Lemma 8.41 it would make {0} → E(1)₀ = {0, 1} a homotopy equivalence of spaces, which it is not (Example 10.4). The two directions of E(1) disagree:

categorical direction 0 1 01 10 0 and 1 are isomorphic spatial direction: E(1)₀ 0 1 no path between 0 and 1
E(1): equivalent as objects, disconnected as points (Intuition 10.5: the homotopy theory hiding inside the category direction is invisible to the space direction).
Examples 10.4 / 10.8 / 10.10 · Three symptoms, one disease

• Equivalences of categories are not sent to equivalences of Segal spaces: ∆⁰ → E(1) (Example 10.4).

• "Fully faithful + essentially surjective ⇒ equivalence" fails, for the same map (Example 10.8, Intuition 10.9).

• E(1) is a Segal space groupoid, but it is not equivalent to ιK for any space K (Example 10.10) — violating the homotopy hypothesis (Intuition 10.11): a groupoid carries no categorical data, so it ought to be a space.

The common source: equivalent objects need not be joined by a path in the object space. Completeness (next section) imposes exactly this.

Exercise · Example 10.14, by the numbers
Show solution
E(1)₀ = {0, 1} (2 elements); E(1)₁ = N(I(1))₁ = {00, 01, 10, 11} (4 elements), and all four are homotopy equivalences since I(1) is a groupoid. So 00 : {0, 1} → {00, 01, 10, 11} misses 01 and 10 — not surjective, hence not an equivalence. E(1) is the minimal incomplete Segal space, and this failure is the exact shape of the next definition.

Section 10 (cont.) · Notes: Def. 10.15 – Thm. 10.44Completeness: gluing the two directions together

Lemma 10.12 / Remark 10.13 / Definition 10.15 · Complete Segal space Every identity idx is a homotopy equivalence (Lemma 10.12), so 00 : W₀ → W₁ factors through Whoequiv (Remark 10.13). A Segal space W is complete if
00 : W₀ → Whoequiv
is a homotopy equivalence. Intuition 10.23: every equivalence in W can be represented by a path in W₀ — the homotopy theory of the object space is the theory of equivalences.
Construction 10.18 / Definition 10.20 · From paths to equivalences Lifting 00 against the Kan fibration Whoequiv → W₀ × W₀ (Lemma 10.17, Remark 10.16) yields a map P : Map(D¹, W₀) → Whoequiv — it takes a path γ from x to y in the object space and produces a homotopy equivalence x → y (Intuition 10.19). Its fiber over (x, y) is the comparison map
Px,y : PathW₀(x, y) → hoequivW(x, y).
Theorems 10.24 / 10.27 / 10.44 · What completeness buys

Equivalence = ff + es (Theorem 10.24): for functors of complete Segal spaces, equivalence is equivalent to fully faithful and essentially surjective — Example 10.8 is repaired.

Homotopy hypothesis (Example 10.26, Proposition 10.27): ιK is always complete, and a complete Segal space is a groupoid iff it is equivalent to ιW₀ — groupoids are spaces, as Intuition 10.11 demanded.

Categories embed completely (Theorem 10.44): the Rezk nerve N(C) — which, unlike disc(NC), puts the isomorphisms of C into the object space as paths (Exercises 10.12, 10.13; Remark 10.28) — is a complete Segal space, with mapNC(x, y) = HomC(x, y), hoequivNC(x, y) = HomC(x, y), Ho(N(C)) ≅ C (Prop. 10.42), and Px,y the identity. Example 10.4 is repaired.

Rzk counterpart · 10-rezk-types: completeness is Rezk-completeness RS17 §10.2: since idx is always an isomorphism, path induction defines
idtoiso : (x =A y) → (x ≅A y)
and a Segal type A is a Rezk type if idtoiso is an equivalence (RS17 Definition 10.6). Line up the two sides: the identity type x = y plays the role of PathW₀(x, y), the type of isomorphisms x ≅ y plays the role of hoequivW(x, y), and idtoiso is exactly the comparison map Px,y of Construction 10.18 — so Rezk-completeness is one of the five equivalent forms of completeness in Proposition 10.22 — the local one: every Px,y is a homotopy equivalence — stated inside the theory. And RS17 Proposition A.25 closes the loop semantically: a Segal space is a Rezk space iff XE□∆⁰ → X is a trivial Reedy fibration — another of the equivalent forms, with E(1) as the probing shape. Segal types are Segal spaces; Rezk types are complete Segal spaces: the model this whole library has been building.