Lemma 69.17.2. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent sheaf on $Y$. Let $\mathcal{I}$ be a quasi-coherent sheaf of ideals on $X$. If $f$ is affine then $\mathcal{I}f_*\mathcal{F} = f_*(f^{-1}\mathcal{I}\mathcal{F})$ (with notation as explained in the proof).

Proof. The notation means the following. Since $f^{-1}$ is an exact functor we see that $f^{-1}\mathcal{I}$ is a sheaf of ideals of $f^{-1}\mathcal{O}_ X$. Via the map $f^\sharp : f^{-1}\mathcal{O}_ X \to \mathcal{O}_ Y$ on $Y_{\acute{e}tale}$ this acts on $\mathcal{F}$. Then $f^{-1}\mathcal{I}\mathcal{F}$ is the subsheaf generated by sums of local sections of the form $as$ where $a$ is a local section of $f^{-1}\mathcal{I}$ and $s$ is a local section of $\mathcal{F}$. It is a quasi-coherent $\mathcal{O}_ Y$-submodule of $\mathcal{F}$ because it is also the image of a natural map $f^*\mathcal{I} \otimes _{\mathcal{O}_ Y} \mathcal{F} \to \mathcal{F}$.

Having said this the proof is straightforward. Namely, the question is étale local on $X$ and hence we may assume $X$ is an affine scheme. In this case the result is a consequence of the corresponding result for schemes, see Cohomology of Schemes, Lemma 30.13.2. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).