Kerodon

Proof. Let $\operatorname{\mathcal{C}}^{+}$ denote the category obtained from $\operatorname{\mathcal{C}}$ by formally adjoining a new identity morphism $\operatorname{id}_{X}^{+}$ for each object $X \in \operatorname{\mathcal{C}}$. More precisely, the category $\operatorname{\mathcal{C}}^{+}$ is defined as follows:

• The objects of $\operatorname{\mathcal{C}}^{+}$ are the objects of $\operatorname{\mathcal{C}}$.

• For every pair of objects $X,Y \in \operatorname{\mathcal{C}}^{+}$, we have

  \[ \operatorname{Hom}_{\operatorname{\mathcal{C}}^{+}}( X,Y ) = \begin{cases} \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) & \text{ if } X \neq Y \\ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \coprod \{ \operatorname{id}^{+}_{X} \} & \text{ if } X = Y. \end{cases} \]

• If $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ are morphisms in $\operatorname{\mathcal{C}}^{+}$, then the composition $g \circ f$ is equal to $g$ if $f = \operatorname{id}_{Y}^{+}$, to the morphism $f$ if $g = \operatorname{id}_{Y}^{+}$, and is otherwise given by the composition law for morphisms in $\operatorname{\mathcal{C}}$.

Note that the collection of non-identity morphisms in $\operatorname{\mathcal{C}}^{+}$ is closed under composition, so that the nerve $\operatorname{N}_{\bullet }( \operatorname{\mathcal{C}}^{+} )$ is a braced simplicial set (Exercise 3.3.1.2). Unwinding the definitions, we see that the semisimplicial subset $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}^{+})^{\mathrm{nd}} \subseteq \operatorname{N}_{\bullet }( \operatorname{\mathcal{C}}^{+} )$ can be identified with the $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ (as a semisimplicial set). Using Corollary 3.3.1.11, we obtain a canonical isomorphism of simplicial sets $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})^{+} \simeq \operatorname{N}_{\bullet }( \operatorname{\mathcal{C}}^{+} )$. Under this isomorphism, the counit map $v_{ \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) }$ is induced by the functor $F: \operatorname{\mathcal{C}}^{+} \rightarrow \operatorname{\mathcal{C}}$ which is the identity on objects, and carries each morphism $f \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \subseteq \operatorname{Hom}_{\operatorname{\mathcal{C}}^{+}}( X, Y)$ to itself.

Let $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}^{+}$ be the functor which is the identity on objects, and which carries a morphism $f \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ to the morphism

this functor is well-defined by virtue of our assumption that the collection of non-identity morphisms of $\operatorname{\mathcal{C}}$ is closed under composition. We will complete the proof by showing that the induced map $\operatorname{N}_{\bullet }(G): \operatorname{N}_{\bullet }( \operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }( \operatorname{\mathcal{C}}^{+} )$ is a simplicial homotopy inverse of $\operatorname{N}_{\bullet }(F) = v_{ \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})}$. One direction is clear: the composition $\operatorname{\mathcal{C}}\xrightarrow {G} \operatorname{\mathcal{C}}^{+} \xrightarrow {F} \operatorname{\mathcal{C}}$ is the identity functor $\operatorname{id}_{\operatorname{\mathcal{C}}}$, so $\operatorname{N}_{\bullet }(F) \circ \operatorname{N}_{\bullet }(G)$ is equal to the identity. The composition $\operatorname{\mathcal{C}}^{+} \xrightarrow {F} \operatorname{\mathcal{C}}\xrightarrow {G} \operatorname{\mathcal{C}}^{+}$ is not the identity functor on $\operatorname{\mathcal{C}}^{+}$: for each object $X \in \operatorname{\mathcal{C}}$, it carries the morphism $\operatorname{id}_{X} \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,X) \subset \operatorname{Hom}_{\operatorname{\mathcal{C}}^{+}}(X,X)$ to the “new” identity morphism $\operatorname{id}_{X}^{+}$. However, there is a natural transformation $\alpha : G \circ F \rightarrow \operatorname{id}_{\operatorname{\mathcal{C}}^{+}}$, given by the construction $(X \in \operatorname{\mathcal{C}}^{+}) \mapsto \operatorname{id}_{X}$. It follows that the map of simplicial sets $\operatorname{N}_{\bullet }(G) \circ \operatorname{N}_{\bullet }(F)$ is homotopic to the identity (Example 3.1.5.7). $\square$