Definition 3.2.6.1. Let $X$ be a simplicial set. We will say that $X$ is *contractible* if the projection map $X \rightarrow \Delta ^{0}$ is a homotopy equivalence (Definition 3.1.6.1). We say that $X$ is *weakly contractible* if the projection map $X \rightarrow \Delta ^{0}$ is a weak homotopy equivalence (Definition 3.1.6.11).

### 3.2.6 Contractibility

In this section, we study the class of *contractible* simplicial sets.

Remark 3.2.6.2. Let $X$ be a simplicial set. If $X$ is contractible, then it is weakly contractible. The converse holds if $X$ is a Kan complex (Proposition 3.1.6.12). Beware that the converse is false in general (Exercise 3.1.6.19).

Example 3.2.6.3. Let $\operatorname{\mathcal{C}}$ be a category. If $\operatorname{\mathcal{C}}$ has an initial object or a final object, then the simplicial set $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ is contractible (this is a special case of Proposition 3.1.6.9). In particular, for every integer $n \geq 0$, the standard simplex $\Delta ^ n$ is contractible.

Remark 3.2.6.4. Let $f: X \rightarrow Y$ be a weak homotopy equivalence of simplicial sets. Then $X$ is weakly contractible if and only if $Y$ is weakly contractible (see Remark 3.1.6.15). If $f$ is a homotopy equivalence, then $X$ is contractible if and only if $Y$ is contractible (see Remark 3.1.6.7).

Example 3.2.6.5. Let $n$ be a positive integer. For $0 \leq i \leq n$, the horn $\Lambda ^{n}_{i}$ is weakly contractible. This follows from Remark 3.2.6.4, since the inclusion map $\Lambda ^{n}_{i} \hookrightarrow \Delta ^ n$ is a weak homotopy equivalence (Proposition 3.1.6.13) and the simplex $\Delta ^ n$ is contractible (Example 3.2.6.3).

Definition 3.2.6.6. Let $f: X \rightarrow Y$ be a morphism of simplicial sets. We will say that $f$ is *nullhomotopic* if there exists a vertex $y \in Y$ for which $f$ is homotopic to the constant morphism $X \rightarrow \{ y\} \hookrightarrow Y$.

Example 3.2.6.7. Let $X$ be a simplicial set, and let $\emptyset $ denote the empty simplicial set. Then there is a unique morphism of simplicial sets $\emptyset \hookrightarrow X$, which is nullhomotopic if and only if $X$ is nonempty (note that, by the convention of Definition 3.2.6.6, the identity map $\emptyset \rightarrow \emptyset $ is *not* considered to be nullhomotopic).

Remark 3.2.6.8. Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be morphisms of simplicial sets. If either $f$ or $g$ is nullhomotopic, then the composition $g \circ f$ is nullhomotopic.

Proposition 3.2.6.9. Let $Y$ be a simplicial set. The following conditions are equivalent:

- $(1)$
Every morphism of simplicial sets $f: X \rightarrow Y$ is nullhomotopic.

- $(2)$
Every morphism of simplicial sets $g: Y \rightarrow Z$ is nullhomotopic.

- $(3)$
The identity morphism $\operatorname{id}_{Y}: Y \rightarrow Y$ is nullhomotopic.

- $(4)$
The simplicial set $Y$ is contractible.

**Proof.**
The implications $(1) \Rightarrow (3)$ and $(2) \Rightarrow (3)$ are immediate, and the reverse implications follow from Remark 3.2.6.8. To see that $(3) \Leftrightarrow (4)$, it suffices to observe that a morphism $y: \Delta ^0 \rightarrow Y$ is homotopy inverse to the projection map $Y \rightarrow \Delta ^0$ if and only if the identity morphism $\operatorname{id}_{Y}$ is homotopic to the constant morphism $Y \twoheadrightarrow \{ y\} \hookrightarrow Y$.
$\square$

Variant 3.2.6.10. A simplicial set $Y$ is weakly contractible if and only if, for every Kan complex $Z$, every morphism $f: Y \rightarrow Z$ is nullhomotopic.

**Proof.**
Without loss of generality, we may assume that $Y$ is nonempty (note that if $Y$ is empty, then $Y$ is a Kan complex but the identity map $\operatorname{id}_{Y}: Y \rightarrow Y$ is not nullhomotopic). Fix a vertex $y \in Y$. By definition, $Y$ is weakly contractible if and only if, for every Kan complex $Z$, the diagonal map $\delta : Z \rightarrow \operatorname{Fun}(Y,Z)$ induces a bijection on connected components. Note that $\delta $ admits a left inverse (given by the evaluation map $\operatorname{Fun}(Y,Z) \rightarrow \operatorname{Fun}( \{ y\} , Z) \simeq Z$), and is therefore automatically injective on connected components. Consequently, $Y$ is weakly contractible if and only if, for every Kan complex $Z$, the map $\delta $ is surjective at the level of connected components: that is, if and only if every morphism $f: Y \rightarrow Z$ is homotopic to a constant map.
$\square$

Proposition 3.2.6.11. Let $X$ be a Kan complex and let $n \geq 0$ be an integer. Then a morphism of simplicial sets $\sigma _0: \operatorname{\partial \Delta }^ n \rightarrow X$ is nullhomotopic if and only if it can be extended to an $n$-simplex $\sigma : \Delta ^ n \rightarrow X$.

**Proof.**
Suppose first that $\sigma _0$ is homotopic to a constant map $\sigma '_{0}: \operatorname{\partial \Delta }^ n \rightarrow \{ x\} \hookrightarrow X$. Since $\sigma '_{0}$ can be extended to a map $\sigma ': \Delta ^ n \rightarrow \{ x\} \hookrightarrow X$, it follows from the homotopy extension lifting property (Remark 3.1.5.3) that $\sigma _0$ can be extended to an $n$-simplex of $X$.

For the converse, assume that $\sigma _0$ admits an extension $\sigma : \Delta ^ n \rightarrow X$. Since the simplex $\Delta ^ n$ is contractible (Example 3.2.6.3), it is weakly contractible (Remark 3.2.6.2), so the morphism $\sigma : \Delta ^ n \rightarrow X$ is nullhomotopic (Variant 3.2.6.10). Applying Remark 3.2.6.8, we deduce that $\sigma _0 = \sigma |_{ \operatorname{\partial \Delta }^ n }$ is also nullhomotopic. $\square$

Proposition 3.2.6.12. Let $f: (X,x) \rightarrow (Y,y)$ be a morphism of pointed simplicial sets. Suppose that $Y$ is a Kan complex. The following conditions are equivalent:

- $(1)$
The morphism $f$ is nullhomotopic as an unpointed map. That is, there exists a vertex $z \in Y$ and a homotopy from $f$ to the constant map $\underline{z}: X \rightarrow Y$ taking the value $z$.

- $(2)$
The morphism $f$ is nullhomotopic as a pointed map: that is, there exists a pointed homotopy from $f$ to the constant map $\underline{y}: X \rightarrow Y$.

**Proof.**
The implication $(2) \Rightarrow (1)$ is immediate from the definition. To prove the converse, suppose that there exists a a homotopy $h: \Delta ^1 \times X \rightarrow Y$ satisfying $h|_{ \{ 0\} \times X} = f$ and $h |_{ \{ 1\} \times X} = \underline{z}$ for some vertex $z \in Y$. Let $e: y \rightarrow z$ be the edge of $Y$ given by the restriction $h|_{ \Delta ^1 \times \{ x\} }$ and let $\sigma = s_0(e)$ denote the degenerate $2$-simplex of $Y$ depicted in the diagram

Let $\underline{e}: \underline{y} \rightarrow \underline{y}'$ denote the image fo $e$ in $\operatorname{Fun}(X,Y)$. Since $Y$ is a Kan complex, the restriction map $q: \operatorname{Fun}(X,Y) \rightarrow \operatorname{Fun}(\{ x\} ,Y) \simeq Y$ is a Kan fibration (Corollary 3.1.3.3). It follows that the lifting problem

which carries the edge $\operatorname{N}_{\bullet }( \{ 0 < 1 \} ) \subseteq \Delta ^2$ to a pointed homotopy from $f$ to $\underline{y}$. $\square$

Example 3.2.6.13. Let $(X,x)$ be a pointed Kan complex, let $n > 0$ be a positive integer, and let $\sigma : \Delta ^ n / \operatorname{\partial \Delta }^ n \rightarrow (X,x)$ be a morphism of pointed simplicial sets. Then $\sigma $ is nullhomotopic (in the sense of Definition 3.2.6.6) if and only if the pointed homotopy class $[\sigma ]$ is equal to the identity element in the homotopy group $\pi _{n}(X,x)$.

Proposition 3.2.6.14. Let $X$ be a Kan complex. The following conditions are equivalent:

- $(1)$
For every integer $n \geq 0$, every morphism of simplicial sets $\operatorname{\partial \Delta }^ n \rightarrow X$ is nullhomotopic.

- $(2)$
The projection map $X \rightarrow \Delta ^{0}$ is a trivial Kan fibration.

- $(3)$
The Kan complex $X$ is contractible.

- $(4)$
The Kan complex $X$ is nonempty. Moreover, for every integer $n \geq 0$, every morphism of simplicial sets $\Delta ^ n / \operatorname{\partial \Delta }^ n \rightarrow X$ is nullhomotopic.

- $(5)$
The Kan complex $X$ is nonempty and the set $\pi _{n}(X,x)$ has a single element for each vertex $x \in X$ and each $n \geq 0$.

- $(6)$
The Kan complex $X$ is connected. Moreover, there exists a vertex $x \in X$ such that the homotopy groups $\pi _{n}(X,x)$ are trivial for $n \geq 1$.

**Proof.**
The equivalence $(1) \Leftrightarrow (2)$ follows from Proposition 3.2.6.11, the implication $(2) \Rightarrow (3)$ is a special case of Proposition 3.1.6.10, the implication $(3) \Rightarrow (4)$ follows from Proposition 3.2.6.9, the equivalence $(4) \Leftrightarrow (5)$ from Example 3.2.6.13, and the implication $(5) \Rightarrow (6)$ is immediate.

We will complete the proof by showing that $(6)$ implies $(1)$. Assume that $X$ is connected and let $x \in X$ be a vertex for which the homotopy groups $\pi _{n}(X,x)$ vanish for $n \geq 1$. Let $\sigma : \operatorname{\partial \Delta }^ n \rightarrow X$ be a morphism of simplicial sets for some $n \geq 0$; we wish to show that $\sigma $ is nullhomotopic. For $n \leq 1$, this follows immediately from our assumption that $X$ is connected. We may therefore assume without loss of generality that $n \geq 2$. Let $\sigma _0$ denote the restriction $\sigma |_{ \Lambda ^{n}_{n} }$. Since the horn $\Lambda ^{n}_{n}$ is weakly contractible, the morphism $\sigma _0$ is nullhomotopic (Variant 3.2.6.10). Using the connectedness of $X$ we see that $\sigma _0$ is homotopic to the constant morphism $\sigma '_0: \Lambda ^{n}_{n} \rightarrow \{ x\} \hookrightarrow X$. Applying the homotopy extension lifting property (Remark 3.1.5.3), we conclude that $\sigma $ is homotopic to a morphism $\sigma ': \operatorname{\partial \Delta }^ n \rightarrow X$ satisfying $\sigma '|_{\Lambda ^{n}_{n}} = \sigma '_0$. It will therefore suffice to show that $\sigma '$ is nullhomotopic. Note that $\sigma '$ factors as a composition

where $\tau $ carries the base point of $\Delta ^{n-1} / \operatorname{\partial \Delta }^{n-1}$ to the point $x$. Since the homotopy group $\pi _{n-1}(X,x)$ vanishes, Example 3.2.6.13 guarantees that $\tau $ is nullhomotopic, so that $\sigma '$ is also nullhomotopic (Remark 3.2.6.8). $\square$

We establish a relative version of Proposition 3.2.6.14.

Proposition 3.2.6.15. Let $f: X \rightarrow S$ be a Kan fibration of simplicial sets. The following conditions are equivalent:

- $(1)$
The morphism $f$ is a trivial Kan fibration.

- $(2)$
For each vertex $s \in S$, the fiber $X_{s} = \{ s\} \times _{S} X$ is a contractible Kan complex.

- $(3)$
For each vertex $s \in S$, the fiber $X_{s} = \{ s\} \times _{S} X$ is connected. Moreover, for each vertex $x \in X_{s}$, the homotopy groups $\pi _{n}(X_ s, x)$ vanish for $n > 0$.

**Proof.**
The implication $(1) \Rightarrow (2)$ and the equivalence $(2) \Leftrightarrow (3)$ follow by applying Proposition 3.2.6.14 to the fibers of $f$. We will complete the proof by showing that $(2)$ implies $(1)$. Assume that $(2)$ is satisfied; we wish to show that every lifting problem

admits a solution. Let $q: \Delta ^1 \times \Delta ^ n \rightarrow \Delta ^ n$ be the map given on vertices by the formula $q(i,j) = \begin{cases} j & \text{ if } i = 0 \\ n & \text{ if } i = 1. \end{cases}$ Then we can regard $\overline{\sigma } \circ q$ as a homotopy from $\overline{\sigma }: \Delta ^ n \rightarrow S$ to the constant map $\Delta ^ n \rightarrow \{ s\} \subseteq S$, where $s$ denotes the vertex $\overline{\sigma }(n) \in S$. Since $f$ is a Kan fibration, the restriction $(\overline{\sigma } \circ q)|_{ \Delta ^1 \times \operatorname{\partial \Delta }^ n}$ can be lifted to a homotopy $h: \Delta ^1 \times \operatorname{\partial \Delta }^ n \rightarrow X$ from $\sigma _0$ to some map $\sigma '_0: \operatorname{\partial \Delta }^ n \rightarrow X_{s}$ (Remark 3.1.5.3). It follows from assumption $(2)$ that we can extend $\sigma '_0$ to an $n$-simplex $\sigma ': \Delta ^{n} \rightarrow X_{s}$. Invoking our assumption that $f$ is a Kan fibration again, we see that $h$ can be extended to a homotopy $\widetilde{h}: \Delta ^1 \times \Delta ^ n \rightarrow X$ from $\sigma $ to $\sigma '$, where $\sigma : \Delta ^ n \rightarrow X$ is an extension of $\sigma _0$ satisfying $f( \sigma ) = \overline{\sigma }$. $\square$