Proof (0N4G) · ScalingStacks

0N4G

Proof. After Proposition 3.9, we can assume that $B$ is equivalent to $\BTop(n)$ , and so we omit it from the notation and discussion. We will explain the string of canonical equivalences in $\mathcal{V}$ :

$\displaystyle\int_{N}f_{\ast}A$	$\displaystyle\underset{\rm Def~\ref{coend}}{\simeq}$	$\displaystyle\colim_{U\in\disk_{k/N}^{\partial,{\sf or}}}f_{\ast}A(U)$
	$\displaystyle\underset{{\rm Def~}f_{\ast}}{\simeq}$	$\displaystyle\colim_{U\in\disk_{k/N}^{\partial,{\sf or}}}\int_{f^{-1}U}A$
	$\displaystyle\underset{\rm Def~\ref{coend}}{\simeq}$	$\displaystyle\colim_{U\in\disk_{k/N}^{\partial,{\sf or}}}\colim_{V\in\disk_{n/f^{-1}U}}A(V)$
	$\displaystyle\underset{(1)}{\simeq}$	$\displaystyle\colim_{(U,V)\in\disk_{f}}A(V)$
	$\displaystyle\underset{(2)}{\xrightarrow{\simeq}}$	$\displaystyle\colim_{V\in\disk_{n/M}}A(V)$
	$\displaystyle\underset{\rm Def~\ref{coend}}{\simeq}$	$\displaystyle\int_{M}A~.$

The only equivalences that are not definitional are (1) and (2). The equivalence (2) is a direct application of Lemma 3.21, which states that the functor ${\sf ev}_{0}\colon\disk_{f}\to\disk_{n/M}$ is final. Consider the left Kan extension (non-commutative) diagram among $\infty$ -categories:

which exists because $\mathcal{V}$ is presentable. By construction, the functor ${\sf ev}_{1}\colon\disk_{f}\to\disk^{\partial,\sf or}_{k/N}$ is a coCartesian fibration. In particular, for each object $U\in\disk^{\partial,\sf or}_{k/N}$ , the inclusion of the fiber into the over $\infty$ -category

{\sf ev}_{1}^{-1}(U)\longrightarrow(\disk_{f})_{/U}

is final. Therefore, the value of ${\sf LKan}$ on $U\in\disk^{\partial,\sf or}_{k/N}$ is the colimit over the fiber:

\colim_{V\in\disk_{n/f^{-1}U}}A(V)~\underset{\rm Def~\ref{disk-f}}{\simeq}~\colim_{V\in{\sf ev}_{1}^{-1}U}A(V)\xrightarrow{~\simeq~}{\sf LKan}(U)~.

So the colimit of ${\sf LKan}$ is the codomain of (1). The equivalence (1) follows from Proposition 4.3.3.7 of [Lu1], which implies the colimit of a left Kan extension agrees with the colimit because they both satisfy the same universal property.

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