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I'm trying to show the following statement from this answer

$$f_* \mathscr{H}om_X(A,\;B) \cong \mathscr{H}om_Y(f_* A,\; f_* B)$$

where $ f:X\rightarrow Y$ is a map of topological spaces and $ A,B$ are sheaves on $ X $.

I must be having a misunderstanding here, but I can't see how this is true. Suppose $ Y $ is a singleton. Take global sections. The left side is $ \rm Hom_X(A,B) $, the group of morphisms from $ A $ to $ B $. The right side is $\rm Hom(A(X), B(X))$, the group of group homomorphisms from $ A(X) $ to $ B(X) $. Are these two really isomorphic in general? There must be a flaw in my reasoning.

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PeterM
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1 Answers1

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As you correctly observe, there is no such isomorphism in general, just a morphism from the left hand side to the right hand side. There is a correct formula involving sheaf-hom, which is as follows:

$$f_* \mathscr{H}{om}_X(f^* A, B) \cong \mathscr{H}{om}_Y(A, f_* B),$$

for a sheaf $A$ on $Y$ and a sheaf $B$ on $X$. This does follow from the definition of $\mathscr{H}om$ and the usual adjunction between $f^*$ and $f_*$.

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