Kerodon

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 $\sigma : \Delta ^ n \rightarrow X$. The projection map $\Delta ^ n \rightarrow \Delta ^0$ is a homotopy equivalence (Proposition 3.1.6.9). It follows that the induced map $X \simeq \operatorname{Fun}( \Delta ^0, X) \rightarrow \operatorname{Fun}( \Delta ^ n, X)$ is also a homotopy equivalence. In particular, it induces a surjection on connected components, so $\sigma $ is nullhomotopic. Applying Remark 3.2.4.16, we deduce that $\sigma _0 = \sigma |_{ \operatorname{\partial \Delta }^ n }$ is also nullhomotopic. $\square$

Proof. Our proof proceeds by induction on $n$. We first show that $(2)$ implies $(3)$. Applying assumption $(2)$ to the inclusion map $\emptyset \subseteq \Delta ^0$, we deduce that there exists a vertex $x \in X$. It will therefore suffice to show that if $A$ is a simplicial set of dimension $< n$, then every pair of morphisms $f_0, f_1: A \rightarrow X$ are homotopic (in particular, $f_0$ is homotopic to the constant map $A \rightarrow \{ x\} $). This follows by applying $(2)$ to the inclusion map $\operatorname{\partial \Delta }^1 \times A \hookrightarrow \Delta ^1 \times A$ (see Proposition 1.1.3.11).

The implication $(3) \Rightarrow (4)$ follows from Lemma 3.2.4.17. We next show that $(4)$ implies $(1)$. Applying assumption $(4)$ in the case $m = 0$, we conclude that $X$ is nonempty (Example 3.2.4.14). Fix a vertex $x \in X$; we wish to show that every element $\eta \in \pi _{k}(X,x)$ is trivial for $0 \leq k < n$. Choose a $k$-simplex $\sigma : \Delta ^{k} \rightarrow X$ which represents $\eta $, so that $\sigma |_{ \operatorname{\partial \Delta }^ k }$ is the constant morphism taking the value $x$ Then $\sigma $ can be extended to a map $\tau _0: \operatorname{\partial \Delta }^{n+1} \rightarrow X$ for which $\tau _0 |_{ \Lambda ^{k+1}_{0} }$ is the constant morphism taking the value $x$. Assumption $(4)$ guarantees that $\tau _0$ is can be extended to an $(k+1)$-simplex $\tau $ of $X$ (Lemma 3.2.4.17). Let $u: \Delta ^1 \times \Delta ^ k \rightarrow \Delta ^{k+1}$ be the morphism given on vertices by the formula

Then the composition $\Delta ^1 \times \Delta ^{k} \xrightarrow {\rho } \Delta ^{k+1} \xrightarrow {\tau } X$ determines a homotopy from $\sigma $ to the constant map $\Delta ^{k} \rightarrow \{ x\} \hookrightarrow X$, which is constant along the boundary $\operatorname{\partial \Delta }^{k}$.

We now complete the proof by showing that $(1)$ implies $(2)$. Assume that $X$ is $n$-connective, let $B$ be a simplicial set of dimension $\leq n$, let $A \subseteq B$ be a simplicial subset, and suppose we are given a morphism $f_0: A \rightarrow X$; we wish to show that $f_0$ admits an extension $B \rightarrow X$. Using Proposition 1.1.3.13, we can reduce to the case where $B = \Delta ^{m}$ is a standard simplex of dimension $m \leq n$ and $A = \operatorname{\partial \Delta }^{m}$ is its boundary. Since $X$ is nonempty, we may assume that $m > 0$. If $m = 1$, then $X$ is connected (Example 3.2.4.11), and the desired result follows from Proposition 1.1.9.10. We may therefore assume without loss of generality that $m \geq 2$. Let $K$ be the $(m-2)$-skeleton of $\Delta ^{m}$. Since $X$ is $(n-1)$-connective, our inductive hypothesis guarantees that $f_0|_{ K }$ is nullhomotopic. Using Remark 3.1.5.3, we can choose a homotopy $\alpha _0$ from $f_0$ to a morphism $g_0: \operatorname{\partial \Delta }^{m} \rightarrow X$ such that $g_0|_{ K }$ is the constant map taking some value $x \in X$. Since $m \leq n$, our assumption that $X$ is $n$-connective guarantees that the homotopy group $\pi _{m-1}(X,x)$ is trivial. Using the characterization of the group structure on $\pi _{m-1}(X,x)$ given in Theorem 3.2.2.10, we conclude that $g_0$ admits an extension $g: \Delta ^{m} \rightarrow X$. Using Remark 3.1.5.3 again, we can extend $\alpha _0$ to a homotopy from $f$ to $g$, where $f: \Delta ^{m} \rightarrow X$ is some extension of $f_0$. $\square$

Proof of Theorem 3.2.4.3. Let $X$ be a Kan complex. We wish to show that the following conditions are equivalent:

$(1)$

The Kan complex $X$ is contractible: that is, the projection map $X \rightarrow \Delta ^0$ is a homotopy equivalence.

$(2)$

The Kan complex $X$ is $n$-connective for every integer $n \geq 0$.

$(3)$

The projection map $X \rightarrow \Delta ^0$ is a trivial Kan fibration: that is, every morphism of simplicial sets $\operatorname{\partial \Delta }^ n \rightarrow X$ can be extended to an $n$-simplex of $X$.

The implication $(1) \Rightarrow (2)$ follows from Remark 3.2.2.16, and the implication $(3) \Rightarrow (1)$ is a special case of Proposition 3.1.6.10. The equivalence of $(2)$ and $(3)$ follows from Proposition 3.2.4.18. $\square$

Proof. We first show that $(1) \Leftrightarrow (2)$. Using Corollary 3.1.7.2, we can choose an anodyne morphism $X \hookrightarrow X'$, where $X'$ is a Kan complex. By virtue of Remark 3.2.4.21, we can replace $X$ by $X'$ and thereby reduce to proving Proposition 3.2.4.23 under the assumption that $X$ is a Kan complex. In this case, $X$ is weakly contractible if and only if it is contractible (Remark 3.2.4.20). The desired result now follows from Proposition 3.2.4.18.

We now show that $(1) \Leftrightarrow (3)$. Without loss of generality, we may assume that $X$ is nonempty (note that if $X$ is empty, then $X$ is a Kan complex but the identity map $\operatorname{id}_{X}: X \rightarrow X$ is not nullhomotopic). Fix a vertex $x \in X$. By definition, $X$ is weakly contractible if and only if, for every Kan complex $Y$, the diagonal map $\delta : X \rightarrow \operatorname{Fun}(X,Y)$ induces a bijection on connected components. Note that $\delta $ admits a left inverse (given by the evaluation map $\operatorname{Fun}(X,X) \rightarrow \operatorname{Fun}( \{ x\} , X) \simeq Y$), and is therefore automatically injective on connected components. Consequently, $X$ is weakly contractible if and only if, for every Kan complex $Y$, the map $\delta $ is surjective at the level of connected components: that is, if and only if every morphism $f: X \rightarrow Y$ is homotopic to a constant map. $\square$