# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Proposition 7.2.1.3. Let $f: A \rightarrow B$ be a morphism of simplicial sets. Then $f$ is left anodyne if and only if it is a left cofinal monomorphism. Similarly, $f$ is right anodyne if and only if it is a right cofinal monomorphism.

Proof. We will prove the first assertion; the second follows by a similar argument. Assume first that $f$ is left anodyne. Then $f$ is a monomorphism (Remark 4.2.4.4). For every left fibration of simplicial sets $\widetilde{B} \rightarrow B$, the restriction map $\theta : \operatorname{Fun}_{/B}( B, \widetilde{B} ) \rightarrow \operatorname{Fun}_{/B}(A, \widetilde{B} )$ is a pullback of the map

$\operatorname{Fun}(B, \widetilde{B} ) \rightarrow \operatorname{Fun}(B,B) \times _{ \operatorname{Fun}(A,B) } \operatorname{Fun}(A, \widetilde{B} ),$

and is therefore a trivial Kan fibration (Proposition 4.2.5.4). In particular, $u$ is a homotopy equivalence (Proposition 3.1.6.10). Allowing $\widetilde{B}$ to vary, we conclude that $f$ is left cofinal.

We now prove the converse. Assume that $f$ is a left cofinal monomorphism; we wish to show that $f$ is left anodyne. By virtue of Proposition 4.2.4.5, it will suffice to show that every lifting problem

7.8
$$\begin{gathered}\label{equation:cofinal-vs-anodyne} \xymatrix@R =50pt@C=50pt{ A \ar [d]^{f} \ar [r] & X \ar [d]^{q} \\ B \ar@ {-->}[ur] \ar [r]^-{g} & S } \end{gathered}$$

admits a solution, provided that $q$ is a left fibration of simplicial sets. Let us regard the morphism $g$ as fixed, and consider the restriction map

$\theta : \operatorname{Fun}_{/B}( B, X \times _{S} B ) \rightarrow \operatorname{Fun}_{ / B }( A, X \times _{S} B).$

Since $f$ is a monomorphism, the morphism $\theta$ is a left fibration (Proposition 4.2.5.1). Since the target simplicial set $\operatorname{Fun}_{ / B }( A, X \times _{S} B)$ is a Kan complex (Corollary 4.4.2.5), it follows that $\theta$ is a Kan fibration (Corollary 4.4.3.8). Our assumption that $f$ is left cofinal guarantees that $\theta$ is a homotopy equivalence, and therefore a trivial Kan fibration (Proposition 3.2.7.2). In particular, it is surjective at the level of vertices, which guarantees that (7.8) admits a solution. $\square$