Canonical morphism $H^{n}(Y,f_* \mathscr{F}) \to H^{n}(X, \mathscr{F})$ in sheaf cohomology.

Question

Given a scheme morphism $f:X \to Y$ and a sheaf of abelian groups $\mathscr{F}$ on $X$ we can consider an injective resolution of this sheaf by a complex $I^{\bullet}$ namely an exact sequence,

$$0 \to \mathscr{F} \to I^0 \to I^1 \to I^2 \to \cdots $$

with each $I^{n}$ an injective sheaf. Then since $f^{-1}$ is an exact functor between the categories of abelian sheaves on $Y$ and the category of abelian sheaves of $X$ we obtain that $f_* I^{n}$ is also injective for every $n$. Then if we denote by $J^{\bullet}$ an injective resolution of $f_* \mathscr{F}$ we can obtain by basic homological algebra ( dual to Weibel Theorem 2.2.6) a morphism of resolutions $f_* I^{\bullet} \to J^{\bullet}$ where $f_* I^\bullet$ denotes the pushfoward of the complex $I^\bullet$.

Then it is clear that if we take global sections and compute cohomology we obtain the desired canonical homomorphism since it is easy to show that does not depend on the injective resolution taken.

My question is: How can we define this canonical map when we are working with $\mathscr{F}$ a quasi-coherent sheaf on $X$? We will assume that $f_* \mathscr{F}$ is also quasi-coherent (for example if $f$ is quasi-compact and quasi-separated). In this case we need to work with $f^{*}$ insted of $f^{-1}$ and I cannot assure that $f_* I^{n}$ are going to be injective $\mathcal{O}_Y$ modules.

Answer 1

As MooS said, in this particular situation, you can use the category of $\mathcal{O}$-module to construct the morphism you seek. In general this is a good method because it leads to a morphism generally easier to construct and to handle.

But let us see the general method where we won't even assume that we are working with the category of sheaves and that the functor involved have an adjoint. So let $F:\mathcal{A}\rightarrow\mathcal{B}$ and $G:\mathcal{B}\rightarrow\mathcal{C}$ be left exact functor such that for every injective $I\in\mathcal{A}$ the object $F(I)$ is $G$-acyclic (ie $(R^iG)(F(I))=0$ for $i>0$). And let us construct a morphism $(R^iG)(F(X))\rightarrow R^i(GF)(X)$.

This is exactly your situation with $F=f_*$ and $G=\Gamma(Y,.)$ and $GF=\Gamma(X,.)$. If $I$ is an injective quasi-coherent sheaf, then it is flabby and so is $f_*I$, we deduce that $f_*I$ is $\Gamma(Y,.)$-acyclic.

Let $X\in \mathcal A$ be any object and $X\rightarrow I^.$ be an injective resolution. Let $F(X)\rightarrow F(I^.)$ be its image by $F$. We would like to construct an injective resolution $F(X)\rightarrow J^.$ together with a morphism $J^.\rightarrow F(I^.)$. (note that the morphism in your question goes in the wrong way). But we won't be able to do it in such a generality.

However, we can construct a quasi-isomophism $F(I^.)\overset\sim\rightarrow K^.$, an injective resolution $F(X)\rightarrow J^.$ and a morphism $J^.\rightarrow K^.$ inducing the identity on $H^0=F(X)$ (and this construction is unique up to homotopy).

The idea is to construct a bicomplex of injectives : start with the line $$0\rightarrow F(X)\rightarrow F(I^0)\rightarrow F(I^1)\rightarrow ...$$ Now for each object $F(I^i)$ in this line, construct an injective resolution $F(I^i)\rightarrow K^{.,i}$ together with morphism of injective resolution compatible with the differentials of this line. Such a bicomplex exists !

Now the first column is an injective resolution $F(X)\rightarrow J^.$. Seeing $J^.$ as a bicomplex concetrated in one column, you have a map of bicomplex $J^.\rightarrow K^{.,.}$. Taking the total complexes you get $J^.\rightarrow K^.:=\operatorname{Tot}K^{.,.}$. Similarly, seeing $F(I^.)$ as a bicomplex concentrated in one line, you have a morphism of bicomplexes $F(I^.)\rightarrow K^{.,.}$ and taking the total complexes you get $F(I^.)\rightarrow K^.$. This last morphism is a quasi-isomorphism since $F(I^.)\rightarrow K^{.,.}$ was a quasi-isomorphism.

And you are done ! Apply $G$ and you have : $$ R^iG(F(X))=H^i(G(J^.))\rightarrow H^i(G(K^.))\overset{\sim}\leftarrow H^i(GF(I^.))=R^i(GF)(X)$$

In fact, this is how you construct the Grothendieck spectral sequence. The map you seek is in fact one of the boundary map.

Edit : I forgot to explain why $GF(I^.)\rightarrow G(K^.)$ is a quasi-isomorphism. As I said, $F(I^.)\rightarrow K^{.,.}$ is column-wise a quasi-isomorphism. But if we apply $G$, it remains true. This is because we assumed that $F(I)$ are $G$-acyclic. So $GF(I^.)\rightarrow GK^{.,.}$ is column-wise a quasi-isomorphism. Taking total complexes, $GF(I^.)\rightarrow \operatorname{Tot} GK^{.,.}=GK^.$ is a quasi-isomorphism.