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Proposition Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be morphisms of simplicial sets, and suppose that $f$ is left cofinal. Then $g$ is left cofinal if and only if the composite map $g \circ f$ is left cofinal. In particular, the collection of left cofinal morphisms is closed under composition.

Proof. Let $q: \widetilde{C} \rightarrow C$ be a left fibration of simplicial sets, and let

\[ \operatorname{Fun}_{/C}(C, \widetilde{C}) \xrightarrow { g^{\ast } } \operatorname{Fun}_{/C}( B, \widetilde{C} ) \xrightarrow { f^{\ast } } \operatorname{Fun}_{/C}( A, \widetilde{C} ) \]

be the morphisms given by precomposition with $g$ and $f$. Our assumption that $f$ is left cofinal guarantees that $f^{\ast }$ is a homotopy equivalence. It follows that $g^{\ast }$ is a homotopy equivalence if and only if $f^{\ast } \circ g^{\ast }$ is a homotopy equivalence (Remark $\square$