# Kerodon

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Proposition 6.3.6.8. Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be universally localizing morphisms of simplicial sets. Then the composition $(g \circ f): X \rightarrow Z$ is universally localizing.

Proof. Suppose we are given a morphism of simplicial sets $Z' \rightarrow Z$. Set $X' = Z' \times _{Z} X$, and let $W$ be the collection of those edges $w$ of $X'$ having degenerate image in $Z'$. We will show that the projection map $\pi : X' \rightarrow Z'$ exhibits $Z'$ as a localization of $X'$ with respect to $W$. Set $Y' = Z' \times _{Z} Y$, so that $\pi$ factors as a composition $X' \xrightarrow {f'} Y' \xrightarrow {g'} Z'$. It follows from Proposition 6.3.6.7 (and Remark 6.3.6.6) that $f'$ is a surjection of simplicial sets. In particular, the image $f'(W)$ is the collection of all edges $u$ of $Y'$ having the property that $g'(u)$ is a degenerate edge of $Z'$.

Let $W_0 \subseteq W$ be the collection of those edges $w$ of $X'$ for which $f'(w)$ is a degenerate edge of $Y'$. Applying Proposition 6.3.6.2, we conclude that $f'$ exhibits $Y'$ as a localization of $X'$ with respect to $W_0$, and that $g'$ exhibits $Z'$ as a localization of $Y'$ with respect to $f'(W)$. Applying Proposition 6.3.1.21, we conclude that $\pi = g' \circ f'$ exhibits $Z'$ as the localization of $X'$ with respect to $W_0 \cup W = W$, as desired. $\square$