Section 2.1 · Notes: Def. 2.1 – Lemma 2.5The simplex category ∆, chain notation, and unique decomposition
• 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).
Pick n and i; see the chain notation and element-by-element behavior of dⁱ : [n] → [n+1] and sⁱ : [n] → [n−1].
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).
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.
Write down explicitly the maps d⁰, d¹, d² : [1] → [2] and s⁰ : [1] → [0] using the notation from Notation 2.2.
Show solution
Express the maps 001, 011 ∈ Hom∆([2], [1]) in terms of the generating morphisms sⁱ (write your answer in the form s0).
Show solution
Section 2.1 (cont.) · Notes: Def. 2.7 – Rem. 2.11Simplicial sets and the maps dᵢ, sᵢ
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
• 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)
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}, ...
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.
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.
Compute the following face and degeneracy maps in D²:
Show solution
Notes: Def. 2.12, Rem. 2.18Degenerate and non-degenerate simplices
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.
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.
For each of the following simplices in D²₂, decide whether it is degenerate or non-degenerate:
Show solution
Section 2.3 · Notes: Def. 2.30 – Ex. 2.44Sub-simplicial sets: boundary ∂Dn, horn Lnᵢ, spine Spn
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.
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).
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}.
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.
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.
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.
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)
Compute the non-degenerate simplices of ∂D³: how many are there in dimensions 0, 1, 2, and 3?
Show solution
Based on computing (L²₀)₁, (L²₁)₁, (L²₂)₁: for each horn, which non-degenerate edge of ∂D² is missing?
Show solution
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
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.
Pushout: horns are two edges glued together (Lemma 2.48)
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.
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
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), ..., (αn,βn)) 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.
Identify the non-degenerate 1-simplices and 2-simplices of D¹ × D¹:
Show solution
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 |
|---|---|---|
| D¹ | Δ¹ | \ t → TOP |
| D² | Δ² | \ (t , s) → s ≤ t |
| D³ | 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, L³₂ | Λ³₁, Λ³₂ | inner horns; union of all faces except face k |
| D¹ × D¹ | Δ¹×Δ¹ | shape-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.
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.
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, 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.
On Δ², which edge does each of the three disjuncts in the condition of ∂Δ² correspond to? (Write your answers as chains.)
Show solution
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
• 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:
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.
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.
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
Let σ = (f₁, f₂, f₃) ∈ NC₃ (equivalently, (g, h, k) ∈ NG₃ for a group G). Match each face with its value:
Show solution
Let C be the category x → y → z with the composite g ∘ f. Compute:
Show solution
Section 3.1 (cont.) · Notes: Def. 3.5 – Cor. 3.18The Segal condition and Theorem 3.15
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.
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.
Proposition 3.9 / Remark 3.11: the infinitely many conditions can be packaged into a single one — i∗₂ : 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.
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 = 02∗µ₂(f, g). The category axioms are then read off from well-chosen simplices:
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 = 03∗µ₃(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 = 02∗µ₂(00∗(x), f) = f. Similarly 011∗(f) = s₁(f) gives idy ∘ f = f. (In the notes' proof: f ∘ idx = 02∗µ₂(00∗(x), f) = f and idy ∘ f = 02∗µ₂(f, 00∗(y)) = 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.
-- 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).
In the tetrahedron µ₃(f, g, h), match each face with what it witnesses:
Show solution
Section 3.2 · Notes: Def. 3.19 – Prop. 3.36From algebraic topology: homotopy, CW-complexes, the topological simplex, Sing
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.
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).
|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
Let S be a discrete topological space. What is Sing(S)n, and what kind of simplicial set is Sing(S)?
Show solution
Section 3.2 (cont.) · Notes: Def. 3.37 – Rem. 3.46Kan complexes, π₀, and the homotopy theorem
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.
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.
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)):
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.
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
Compute the number of path-components:
Show solution
Section 4 · Notes: Def. 4.2 – Lemma 4.14Two directions: simplicial spaces and the bisimplicial grid
• 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.
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).
| sSet | categorical direction (disc) | spatial direction (const) |
|---|---|---|
| Dn | ∆n := disc(Dn) | Dl := const(Dl) |
| ∂Dn | ∂∆n := disc(∂Dn) | ∂Dl := const(∂Dl) |
| Lnk | Λnk := disc(Lnk) | — |
| Spn | Spn := 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.
Compute the four sets (disc D¹)nk and compare them with (const D¹)nk. Enter the cardinalities:
Show solution
Identify the level sets of the representables:
Show solution
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.
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.
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:
• 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.
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).
Decide for each:
Show solution
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.
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.
Section 7 · Notes: Def. 7.1 – Rem. 7.16Uniqueness vs. contractibility
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:
(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).
In Proposition 7.12(3), unpack the cases n = 0 and n = 1, using ∂D⁰ = ∅ and ∂D¹ ≅ D⁰ ⨿ D⁰:
Show solution
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
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.
Left: the world of Theorem 3.15 — the Segal condition on a simplicial set makes the fiber of i∗₂ 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).
the fiber over (f, g) is one point
choices of g∘f are homotopic in map_T(x, z)
Section 8 (cont.) · Notes: Def. 8.31 – Prop. 8.45Identities, the homotopy category, and the Rzk dictionary
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.
-- 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
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:
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).
• 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
• 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.
-- 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)
• 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:
• 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.
Show solution
Section 10 (cont.) · Notes: Def. 10.15 – Thm. 10.44Completeness: gluing the two directions together
• 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.