# How to construct base change of cohomology map $H^n(Y,\mathcal{F})\to H^n(X,f^*\mathcal{F})$?

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. 