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I'm trying to show that the tensor product of sheaves commutes with inverse image. I've reduced the problem to the following isomorphism

$$f_*\mathscr{H}om_X(f^*\mathcal{N},\mathcal{P}) \cong \mathscr{H}om_Y(\mathcal{N},f_*\mathcal{P})$$

where $ f:X\rightarrow Y$ is a morphism of ringed spaces, $\mathcal{N}$ is a $\mathcal{O}_Y$ module, and $\mathcal{P} $ is a $ \mathcal{O}_X$ module.

I'm trying to prove this via the adjunction of $f^* $ and $f_ *$, but I'm unable to. Can someone guide me through the steps involved in constructing this isomorphism?

KReiser
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PeterM
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1 Answers1

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$\def\HH{{\mathcal{H}om}}\def\Hom{{\operatorname{Hom}}}\def\N{{\mathcal N}}\def\P{{\mathcal P}}$This comes from checking the natural isomorphism from adjunction on each open set and using naturality to see it induces a morphism of sheaves not just sets.

Explicitly, take an open set $U \subset Y$. Then

$$ f_*\HH_X(f^*\N,\P)(U) = \Hom_{f^{-1}(U)}\left(f^*\N|_{f^{-1}(U)},\P|_{f^{-1}(U)}\right) \enspace \enspace \enspace (*) $$

but $f^*\N|_{f^{-1}(U)} = \left(f|_{f^{-1}(U)}\right)^*\N|_U$. Then by the adjunction, (*) is naturally isomorphic to

$$ \Hom_U\left(\N|_U, f_*\P|_U\right) = \HH_Y(\N,f_*\P)(U). $$

Naturality of this isomorphism tells us that this map commutes with the restrictions and so it is a map of sheaves and so we get an isomorphism.

Elías Guisado Villalgordo
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Dori Bejleri
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  • Which $f^\ast$ is this about? The tensored up one or the naive one? –  Nov 08 '15 at 14:06
  • Also, how exactly does the adjunction yield the isomorphism you mention? I only see how we get an iso $\mathsf{Hom}\left((f|{f^{-1}U})^{\ast}G|{U},F|{f^{-1}U}\right) \cong\mathsf{Hom}\left(G|{U},(f|{f^{-1}U}){\ast}(F|_{f^{-1}U})\right)$ –  Nov 08 '15 at 14:17
  • @Exterior $(f_{|f^{-1}U})*(F{|f^{-1}U})$ is a sheaf on $U$, and explicitly, for open $V\subseteq U$, $(f_{|f^{-1}U})*(F{|f^{-1}U})(V)=F_{|f^{-1}U}(f^{-1}V)=F(f^{-1}V)=(f_*F)_{|U}(V).$ – Lukas Nov 27 '15 at 13:20