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Let $(\rho_{\lambda})_{\lambda}$ a family of injective sheaves on the space $X$ (therefore injective objects on the category of abelian sheaves). Using universal property of product it's easy to see that $\prod_{\lambda} \rho_{\lambda} $ is also an injective sheaf.

My question is why then for the sheaf cohomology holds the isomorphism

$$ H^r(X, \prod_{\lambda} \rho_{\lambda}) \cong \prod_{\lambda} H^r (X,\rho_{\lambda})$$

Ideas:

I know that the sheaf cohomogies induced by injective and acyclical resolutions are identical, but I don't see why this fact may inply the isomorphism above.

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    $\begingroup$ Both sides are obviously zero by the construction of a right derived functor. $\endgroup$
    – MooS
    Mar 9, 2018 at 19:44

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Since injective sheaves are acyclic for global sections, the statement is trivial for $r > 0.$ So all you're claiming is that global sections of the product is the product of global sections. There are various ways to see this -- for example, you could check that the presheaf (sectionwise) product is actually a sheaf, or you could observe that global sections is representable.

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    $\begingroup$ Or the global section functor is right adjoint to the "locally constant sheaf"-functor, hence it commutes with all limits, in particular products. $\endgroup$
    – MooS
    Mar 9, 2018 at 20:11

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