ScalingStacks

Proof. Take any $N\in\qc(Y)$ and $M\in\qc(X)$. First note that for any quasi-coherent sheaf $K\in\qc(Z)$, there is an equivalence $K\simeq\lim_{g\in\it{Aff}/Z}g_{*}g^{*}K$. Thus we calculate

$$f_{+}M\simeq\lim_{\it{Aff}/Y}g_{*}g^{*}f_{+}M\simeq\lim_{\it{Aff}/Y}g_{*}f^{\prime}_{*}g^{\prime*}M\simeq\lim_{\it{Aff}/Y}g_{*}f^{\prime}_{*}g^{\prime*}M\simeq\lim_{\it{Aff}/Y}f_{*}g^{\prime}_{*}g^{\prime*}M\simeq f_{*}(\lim_{\it{Aff}/Y}g^{\prime}_{*}g^{\prime*}M).$$

To prove the lemma, it thus suffices to show that the natural map $M\rightarrow\lim g^{\prime}_{*}g^{\prime*}M$ is an equivalence.

This is a consequence of the stronger claim that the functor $\qc(X)\rightarrow\lim_{\it{Aff}/Y}\qc({X\times_{Y}B})$ is an equivalence. Since the functor $\qc(-)$ takes all colimits of stacks to limits, it therefore suffices to show that the natural map $X\rightarrow\lim_{\it{Aff}/Y}(X\times_{Y}B)$ is an equivalence. This limit can be calculated by picking an affine cover $U\rightarrow Y$, and realizing $Y$ as the geometric realization of the usual simplicial object $U_{*}\to Y$. Finally, since geometric realization commutes with fiber products we are done. ∎