Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 2.2.8.3. Let $\operatorname{\mathcal{C}}$ be a $2$-category and let $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ be the ordinary category of Construction 2.2.8.2, regarded as a $2$-category having only identity $2$-morphisms. Then there is a unique functor of $2$-categories $F: \operatorname{\mathcal{C}}\rightarrow \mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ with the following properties:

  • The functor $F$ carries each object of $\operatorname{\mathcal{C}}$ to itself (regarded as an object of $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$).

  • The functor $F$ carries each $1$-morphism $u: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ to the connected component of $u$, regarded as a vertex of the nerve $\operatorname{N}_{\bullet }( \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y) )$.

Moreover, the functor $F$ exhibits $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ as a coarse homotopy category of $\operatorname{\mathcal{C}}$, in the sense of Definition 2.2.8.1.

Proof. The existence of $F$ follows from Example 2.2.4.14. Let $\operatorname{\mathcal{E}}$ be an ordinary category, and suppose we are given a functor of $2$-categories $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}$. We wish to show that there is a unique functor of ordinary categories $\overline{G}: \mathrm{h} \mathit{\operatorname{\mathcal{C}}} \rightarrow \operatorname{\mathcal{E}}$ satisfying $G = \overline{G} \circ F$. The uniqueness is clear (since the functor $F$ is surjective on objects and on $1$-morphisms). To prove existence, we define $\overline{G}$ on objects by the formula $\overline{G}(X) = G(X)$ and on morphism by using the map of simplicial sets

\[ \operatorname{N}_{\bullet }( \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y) ) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{E}}}( G(X), G(Y) ) \]

and passing to connected components. $\square$