Cohomology of a finite disjoint union

Write $j_a:X_a\hookrightarrow X$ for the inclusions. For every open $V\subseteq X$ one has $V={⨆}_a(V\cap X_a)$ with each $V\cap X_a$ open in $X_a$, and the functor $\rho :\mathcal{F}\mapsto (\mathcal{F}{\mathrm{\mid }}_{X_a}{)}_a$ from $\mathrm{A}\mathrm{b}(X)$ to ${\prod }_a\mathrm{A}\mathrm{b}(X_a)$ has the functor ${\prod }_a(j_a{)}_{\ast }$ as an inverse up to natural isomorphism: the stalk of $(j_a{)}_{\ast }(\mathcal{F}{\mathrm{\mid }}_{X_a})$ at $x\in X_b$ is $(\mathcal{F}{\mathrm{\mid }}_{X_a}{)}_x$ if $b=a$ and $0$ if $b\mathrm{\ne }a$, the second case because $X\setminus X_a$ is open and contains $x$, so the colimit defining the stalk is taken over open sets disjoint from $X_a$; consequently $\rho {\prod }_a(j_a{)}_{\ast }(F_a)=(F_a{)}_a$ and ${\prod }_a(j_a{)}_{\ast }\rho (\mathcal{F})\cong \mathcal{F}$ stalkwise. Hence $\rho$ is an equivalence of categories, and in particular $\mathrm{\Gamma }(X,\mathcal{F})=\mathcal{F}(X)={\prod }_a{\mathcal{F}}_a(X_a)$: a section over $X$ is the same as a compatible family of sections over the $X_a$, by the sheaf axiom applied to the open cover $\{X_a\}$, and by [F2] sections of the product of the direct images over $X$ are the product of the groups ${\mathcal{F}}_a(X_a)$.

Because $j_a^{-1}$ has the exact left adjoint $j_{a!}$ by [F1], the functor $j_a^{-1}$ preserves injective objects: for an injective $I\in \mathrm{A}\mathrm{b}(X)$ and a monomorphism $A\rightarrowtail B$ in $\mathrm{A}\mathrm{b}(X_a)$, the adjunction identifies $\mathrm{Hom}(A,I{\mathrm{\mid }}_{X_a})\cong \mathrm{Hom}(j_{a!}A,I)$ and $\mathrm{Hom}(B,I{\mathrm{\mid }}_{X_a})\cong \mathrm{Hom}(j_{a!}B,I)$, the map $j_{a!}A\to j_{a!}B$ is a monomorphism by exactness of $j_{a!}$, and $I$ injective by [F6] makes the induced map on Hom groups surjective; hence $I{\mathrm{\mid }}_{X_a}$ is injective. Also $j_a^{-1}$ is exact, since exactness of sheaves can be tested on stalks (A sequence of abelian sheaves is exact exactly when it is exact on every stalk) and the stalk of a restriction is the corresponding stalk. Applying this to the resolution $0\to \mathcal{F}\to I^{\bullet }$ shows that $0\to \mathcal{F}{\mathrm{\mid }}_{X_a}\to I^{\bullet }{\mathrm{\mid }}_{X_a}$ is a resolution of $\mathcal{F}{\mathrm{\mid }}_{X_a}$ by injective sheaves on $X_a$, so its cohomology computes $H^{\bullet }(X_a,\mathcal{F}{\mathrm{\mid }}_{X_a})$ by [F5].

Taking global sections in [step 1.1] degree by degree gives an isomorphism of cochain complexes $\mathrm{\Gamma }(X,I^{\bullet })\cong {\prod }_a\mathrm{\Gamma }(X_a,I^{\bullet }{\mathrm{\mid }}_{X_a})$, since the $q$-th term of the right-hand side is ${\prod }_a\mathrm{\Gamma }(X_a,I^q{\mathrm{\mid }}_{X_a})=\mathrm{\Gamma }(X,I^q)$ by the section computation of [step 1.1]. For a finite product of cochain complexes the cohomology of the product is the product of the cohomologies, because the kernel and image of a componentwise differential are the products of the component kernels and images, and quotient by the product of images gives the product of the quotients (only finitely many representatives are needed); hence, using that $H^q(X,\mathcal{F})=R_I^q\mathrm{\Gamma }(X,\mathcal{F})=H^q(\mathrm{\Gamma }(X,I^{\bullet }))$ by [F3], $H^q(X,\mathcal{F})\cong {\prod }_a{H}^q(\mathrm{\Gamma }(X_a,I^{\bullet }{\mathrm{\mid }}_{X_a}))\cong {\prod }_a{H}^q(X_a,\mathcal{F}{\mathrm{\mid }}_{X_a})$, the last step by [step 1.2]. Naturality in $\mathcal{F}$ follows from the functoriality of the supplied resolutions and of the equivalence in [step 1.1]. This proves the theorem; in degree zero it specializes to $H^0(X,\mathcal{F})\cong {\prod }_a\mathrm{\Gamma }(X_a,\mathcal{F}{\mathrm{\mid }}_{X_a})=\mathrm{\Gamma }(X,\mathcal{F})$ by [F7]. ∎