# Kerodon

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

Corollary 4.3.5.18. Let $A$ and $B$ be simplicial sets, and let $u: A \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{A}})$ and $v: B \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{B}})$ be morphisms which exhibit $\operatorname{\mathcal{A}}$ and $\operatorname{\mathcal{B}}$ as the homotopy categories of $A$ and $B$, respectively. Then the composite map

$A \star B \xrightarrow { u \star v } \operatorname{N}_{\bullet }(\operatorname{\mathcal{A}}) \star \operatorname{N}_{\bullet }(\operatorname{\mathcal{B}}) \simeq \operatorname{N}_{\bullet }( \operatorname{\mathcal{A}}\star \operatorname{\mathcal{B}})$

exhibits $\operatorname{\mathcal{A}}\star \operatorname{\mathcal{B}}$ as the homotopy category of $A \star B$.

Proof. Let $\operatorname{\mathcal{C}}$ be a category, and suppose we are given a map of simplicial sets $f: A \star B \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. Applying Corollary 4.3.5.17 to the morphism $f|_{A}$, we deduce that $f$ factors uniquely as a composition

$A \star B \xrightarrow {u \star \operatorname{id}} \operatorname{N}_{\bullet }(\operatorname{\mathcal{A}}) \star B \xrightarrow {f'} \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}).$

Similarly, $f'$ factors uniquely as a composition

$\operatorname{N}_{\bullet }(\operatorname{\mathcal{A}}) \star B \xrightarrow { \operatorname{id}\star v} \operatorname{N}_{\bullet }(\operatorname{\mathcal{A}}) \star \operatorname{N}_{\bullet }(\operatorname{\mathcal{B}}) \xrightarrow {f''} \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}).$

Combining these observations (together with Example 4.3.3.22 and Proposition 1.2.2.1), we conclude that $f$ factors uniquely as a composition

$A \star B \xrightarrow {u \star v} \operatorname{N}_{\bullet }(\operatorname{\mathcal{A}}) \star \operatorname{N}_{\bullet }(\operatorname{\mathcal{B}}) \simeq \operatorname{N}_{\bullet }( \operatorname{\mathcal{A}}\star \operatorname{\mathcal{B}}) \xrightarrow { \operatorname{N}_{\bullet }(F) } \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$

for some functor $F: \operatorname{\mathcal{A}}\star \operatorname{\mathcal{B}}\rightarrow \operatorname{\mathcal{C}}$. $\square$