Finite Proper Maps Preserve Sheaf Cohomology

The direct image $f_{\ast }\mathcal{F}$ is a sheaf: for an open cover of an open $U\subseteq Y$, the inverse-image opens cover $f^{-1}U$; compatible $\mathcal{F}$-sections glue uniquely by [F6]. Now fix $y\in Y$ and list its finite fiber as $f^{-1}(y)=\{x_1,\dots ,x_n\}$. There is a canonical map $(f_{\ast }\mathcal{F}{)}_y\to {\prod }_{j=1}^n{\mathcal{F}}_{x_j}$ sending a germ of a section near $f^{-1}(y)$ to its germs at these points. It is surjective: represent any finite tuple of germs on neighborhoods $U_j$ of $x_j$, shrink these finitely many neighborhoods to pairwise disjoint open neighborhoods $V_j\subseteq U_j$, and put $V={\bigcup }_j{V}_j$. Since $f$ is closed, $Y\setminus f(X\setminus V)$ is an open neighborhood $W$ of $y$ and $f^{-1}(W)\subseteq V$. The chosen sections restrict to a section on the disjoint open cover $f^{-1}(W)={\bigcup }_j(f^{-1}(W)\cap V_j)$, so they glue and realize the given tuple. The map is injective: if a representative on $f^{-1}(U)$ has zero germ at each $x_j$, choose neighborhoods $O_j\subseteq f^{-1}(U)$ on which it vanishes, shrink to disjoint $V_j\subseteq O_j$, and replace $U$ by the neighborhood $U\setminus f(X\setminus {\bigcup }_j{V}_j)$. The representative then vanishes on that full inverse image, so its direct-image germ is zero. If the fiber is empty, $Y\setminus f(X)$ is a neighborhood on which the direct image is zero, giving the same formula with an empty product. Thus $(f_{\ast }\mathcal{F}{)}_y\cong {\prod }_{x\in f^{-1}(y)}{\mathcal{F}}_x$.

Apply the stalk formula of step 1.1 to a short exact sequence of sheaves on $X$. On each $y$ its pushed stalk sequence is a finite product of the exact stalk sequences at points of $f^{-1}(y)$; finite products of abelian groups preserve exactness. By [F3], direct image is therefore an exact functor on sheaves of abelian groups. It also sends a flasque sheaf $\mathcal{G}$ to a flasque sheaf: if $U^{\mathrm{\prime }}\subseteq U$ are open in $Y$, the restriction $(f_{\ast }\mathcal{G})(U)\to (f_{\ast }\mathcal{G})(U^{\mathrm{\prime }})$ is the restriction $\mathcal{G}(f^{-1}U)\to \mathcal{G}(f^{-1}{U}^{\mathrm{\prime }})$, which is surjective because $f^{-1}{U}^{\mathrm{\prime }}\subseteq f^{-1}U$ are open and $\mathcal{G}$ is flasque [F4].

Let $\mathcal{F}\to C^{\bullet }(\mathcal{F})$ be the Godement flasque resolution from [F1]. Exactness of $f_{\ast }$ from step 2.1 makes $f_{\ast }\mathcal{F}\to f_{\ast }{C}^{\bullet }(\mathcal{F})$ a resolution, and step 2.1 makes every term flasque. By [F5], this is a cohomology-acyclic resolution on $Y$. Hence $H^q(Y,f_{\ast }\mathcal{F})$ is the cohomology of $\mathrm{\Gamma }(Y,f_{\ast }{C}^{\bullet }(\mathcal{F}))$. The direct-image definition [F2] identifies each term $\mathrm{\Gamma }(Y,f_{\ast }{C}^p(\mathcal{F}))$ with $\mathrm{\Gamma }(X,C^p(\mathcal{F}))$, whose cohomology is $H^q(X,\mathcal{F})$ by [F1]. These termwise identifications commute with the resolution maps, giving the asserted isomorphism for all $q\ge 0$.