# Kerodon

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

Lemma 5.1.2.4. Let $q: X \rightarrow S$ be a morphism of simplicial sets, where $S$ is a Kan complex. Then the following conditions are equivalent:

$(1)$

The morphism $q$ is a Kan fibration.

$(2)$

The morphism $q$ is a left fibration.

$(3)$

The morphism $q$ is a right fibration.

Proof. We will show that $(1)$ and $(2)$ are equivalent (the equivalence of $(1)$ and $(3)$ follows from a similar argument). The implication $(1) \Rightarrow (2)$ is contained in Example 4.2.1.5 (and does not require the assumption that $S$ is a Kan complex). Conversely, suppose that $q$ is a left fibration. Then the projection map $X \rightarrow \Delta ^0$ is a left fibration (since it can be written as a composition of $q$ with the Kan fibration $S \rightarrow \Delta ^0$). Invoking Proposition 4.4.2.1, we deduce that $X$ is a Kan complex. To prove that $q$ is a Kan fibration, we must show that for every anodyne morphism of simplicial sets $i: A \hookrightarrow B$, the induced map

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

is surjective on vertices (Remark 3.1.2.6). Note that we have a commutative diagram of simplicial sets

5.1
\begin{equation} \label{diagram:Kan-fibration-lemma} \begin{gathered} \xymatrix@R =50pt@C=50pt{ \operatorname{Fun}(B,X) \ar [r] \ar [d] & \operatorname{Fun}(B,S) \ar [d] \\ \operatorname{Fun}(A,X) \ar [r] & \operatorname{Fun}( A, S), } \end{gathered} \end{equation}

where the vertical maps are trivial Kan fibrations (since $i$ is anodyne and $X$ and $S$ are Kan complexes; see Corollary 3.1.3.6). Invoking Corollary 3.4.1.3, we see that the diagram (5.1) is homotopy Cartesian. It follows that $\theta$ is a weak homotopy equivalence of simplicial sets (Example 3.4.1.5), and is therefore bijective on connected components (Remark 3.1.5.4). Since $\theta$ is a left fibration of simplicial sets (Proposition 4.2.3.1), it is surjective on vertices by virtue of Lemma 5.1.2.3. $\square$