Let $f:X\to Y$ be a morphism of schemes, let $\mathcal{F}$ be an abelian sheaf on $Y$, in some topology $\tau$, safely speaking we assume it is the Zariski topology but it shouldn't really matter.

We further assume $f^*:\mathcal{Sh}(Y)\to \mathcal{Sh}(X)$ is exact. Can we easily deduce a map: $$\DeclareMathOperator{\H}{H} \H^n(Y,\mathcal{F})\to\H^n(X,f^*\mathcal{F})?$$

I would prefer a down to earth map, from the definition of cohomology but not using some Lerry spectral sequence abstract nonsense.

My plan of proof:

Consider an injective resolution $\mathcal{F}\to\mathcal{I}^\bullet$, and by exactness we have an resolution $f^*\mathcal{F}\to f^*\mathcal{I}^\bullet$, and if somehow we have that $f^*$ sends injectives to $\Gamma(X,\cdot)$-acyclic sheaves. Then we can calculate cohomology groups with $f^*\mathcal{I}^\bullet$, i.e. $\H^n(X,f^*\mathcal{F})=\H^n(f^*\mathcal{I}(X))$.

There is a map of complexes of sheaves: $$\mathcal{I}^\bullet\to f_*f^*\mathcal{I}^\bullet$$ apply the global section function we have $$\mathcal{I}^\bullet (Y)\to f_*f^*\mathcal{I}^\bullet (Y)=f^*\mathcal{I}^\bullet (X)$$ which induces a map of cohomology groups $$\H^n(X,f^*\mathcal{F})=\H^n(f^*\mathcal{I}(X))$$ So the problem reduces to do we have that $f^*$ sends injectives to $\Gamma(X,\cdot)$-acyclic sheaves when $f^*$ is exact?

I assume this statement is true since I saw it on J. Milnes' book, "Etale Cohomology", pg 85, in his statement he uses etale topology but the idea should be the same for any topology.enter image description here


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