The Stacks project

Proof. We strongly urge the reader to prove this for themselves. Let $W \in Z_{\acute{e}tale}$ and $s \in (g \circ f)_*\mathcal{F}(W) = \mathcal{F}(X_ W)$. Denote $T \subset X_ W$ the support of $s$; this is a closed subset. Observe that $s$ is a section of $(g \circ f)_!\mathcal{F}$ if and only if $T$ is proper over $W$. We have $f_!\mathcal{F} \subset f_*\mathcal{F}$ and hence $g_!f_!\mathcal{F} \subset g_!f_*\mathcal{F} \subset g_*f_*\mathcal{F}$. On the other hand, $s$ is a section of $g_!f_!\mathcal{F}$ if and only if (a) $T$ is proper over $Y_ W$ and (b) the support $T'$ of $s$ viewed as section of $f_!\mathcal{F}$ is proper over $W$. If (a) holds, then the image of $T$ in $Y_ W$ is closed and since $f_!\mathcal{F} \subset f_*\mathcal{F}$ we see that $T' \subset Y_ W$ is the image of $T$ (details omitted; look at stalks).

The conclusion is that we have to show a closed subset $T \subset X_ W$ is proper over $W$ if and only if $T$ is proper over $Y_ W$ and the image of $T$ in $Y_ W$ is proper over $W$. Let us endow $T$ with the reduced induced closed subscheme structure. If $T$ is proper over $W$, then $T \to Y_ W$ is proper by Morphisms, Lemma 29.41.7 and the image of $T$ in $Y_ W$ is proper over $W$ by Cohomology of Schemes, Lemma 30.26.5. Conversely, if $T$ is proper over $Y_ W$ and the image of $T$ in $Y_ W$ is proper over $W$, then the morphism $T \to W$ is proper as a composition of proper morphisms (here we endow the closed image of $T$ in $Y_ W$ with its reduced induced scheme structure to turn the question into one about morphisms of schemes), see Morphisms, Lemma 29.41.4. $\square$