ScalingStacks

Proof. First, we show that for each $n$ there is an equivalence of $\infty$-categories

$$\Gap([n],\mathrm{Fun}^{\ex}({\mathcal{B}},{\mathcal{A}}))\simeq\mathrm{Fun}^{\ex}({\mathcal{B}},\Gap([n],{\mathcal{A}})).$$

Since $\mathrm{Fun}(-,-)$ is defined simply as the mapping simplicial set [52, 1.2.7.2], we have the equivalence

$$\mathrm{Fun}(\mathrm{N}(\Ar[n]),\mathrm{Fun}({\mathcal{B}},{\mathcal{A}}))\simeq\mathrm{Fun}({\mathcal{B}},\mathrm{Fun}(\mathrm{N}(\Ar[n]),{\mathcal{A}})).$$

Since colimits in functor $\infty$-categories are computed pointwise [52, §5.1.2.3] and the $\infty$-category $\mathrm{Fun}^{\ex}({\mathcal{B}},{\mathcal{A}})$ is the full subcategory of $\mathrm{Fun}({\mathcal{B}},{\mathcal{A}})$ spanned by the exact functors, we have a map

$$\Gap([n],\mathrm{Fun}^{\ex}({\mathcal{B}},{\mathcal{A}}))\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\mathrm{Fun}^{\ex}({\mathcal{B}},\Gap([n],{\mathcal{A}})),$$

and lemma 7.3 implies that it is an equivalence. It is now straightforward to check that these comparison maps assemble into the desired simplicial equivalence. ∎