Kerodon

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. $\square$