# Kerodon

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Proposition 4.3.7.2. Let $f: K \rightarrow X$ and $q: X \rightarrow S$ be morphisms of simplicial sets, let $K_0 \subseteq K$ be a simplicial subset, and set $f_0 = f|_{K_0}$. Then:

• If $q$ is a left fibration, then the induced map

$X_{/f} \rightarrow X_{/f_0} \times _{ S_{ / (q \circ f_0) } } S_{/ (q \circ f)}$

is a Kan fibration.

• If $q$ is a right fibration, then the induced map

$X_{f/} \rightarrow X_{f_0/} \times _{ S_{(q \circ f_0)/ } } S_{(q \circ f)/}$

is a Kan fibration.

Proof. We will prove the first assertion; the proof of the second is similar. Assume that $q$ is a left fibration; we wish to show that the map $X_{/f} \rightarrow X_{/f_0} \times _{ S_{ / (q \circ f_0) } } S_{/ (q \circ f)}$ is a Kan fibration. Equivalently, we wish to show that every lifting problem

$\xymatrix@C =50pt@R=50pt{ A \ar [r] \ar [d] & X_{/f} \ar [d] \\ A' \ar [r] \ar@ {-->}[ur] & X_{/f_0} \times _{ S_{ / (q \circ f_0) } } S_{/ (q \circ f)} }$

admits a solution, provided that the left vertical map $A \rightarrow A'$ is anodyne. Unwinding the definitions, we see that this can be rephrased as a lifting problem

$\xymatrix@C =50pt@R=50pt{ (A \star K) \coprod _{ (A \star K_0) } (A' \star K_0) \ar [r] \ar [d] & X \ar [d]^{q} \\ A' \star K \ar [r] \ar@ {-->}[ur] & S. }$

This problem admits a solution, since the vertical map on the left is left anodyne (Proposition 4.3.7.1) and $q$ is a left fibration. $\square$