Kerodon

Proof. By definition, a morphism $f$ of $\operatorname{\mathcal{C}}$ is an isomorphism if and only if its homotopy class $[f]$ is an isomorphism in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ (Definition 1.4.6.1). By virtue of Remark 5.4.6.4, we can replace $\operatorname{\mathcal{C}}$ by the nerve $\operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} )$ and thereby reduce to the case where $\operatorname{\mathcal{C}}= \operatorname{N}_{\bullet }( \operatorname{\mathcal{C}}' )$ for some category $\operatorname{\mathcal{C}}'$. Let $\sigma $ be a $3$-simplex of $\operatorname{\mathcal{C}}$, corresponding to a triple of morphisms

in $\operatorname{\mathcal{C}}'$, and suppose that $g \circ f$ and $h \circ g$ are isomorphisms. Then $g \circ f$ admits an inverse $u: C \rightarrow A$. It follows that $g \circ (f \circ u) = (g \circ f) \circ u = \operatorname{id}_{C}$, so that $g$ admits a right inverse. A similar argument shows that $g$ also admits a left inverse, and is therefore an isomorphism (Remark 1.4.6.7). Applying the two-out-three property, we deduce that $f$ and $h$ are also isomorphisms. Since the collection of isomorphisms is closed under composition, it also follows that $h \circ g \circ f$ is an isomorphism. $\square$