A detail in the proof that tensor product of sheaves of $\mathcal{O}_X$-modules commutes with pullback

Question

Given a morphism of ringed spaces $f:(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ and $\mathcal{O}_Y$-modules $\mathcal{M}$ and $\mathcal{N}$, here it is proven that $$ f^*(\mathcal{M} \otimes_{\mathcal{O}_Y} \mathcal{N}) \cong f^*\mathcal{M} \otimes_{\mathcal{O}_X} f^*\mathcal{N}. $$ I understand all the proof except for the bit in which it is claimed that

one can check from the definition of $\mathcal{H}om$ and the adjointness of $f^*$ and $f_*$ that $$\tag{1}\label{eq2}f_*\mathcal{H}om_{\mathcal{O}_X}(f^*\mathcal{N},\mathcal{P}) \cong \mathcal{H}om_{\mathcal{O}_Y}(\mathcal{N},f_*\mathcal{P})$$ as $\mathcal{O}_Y$-modules.

Let me explain what I've thought so far: It is easy to verify that in general one has $f_*\mathcal{H}om_{\mathcal{O}_X}(\mathcal{M},\mathcal{N})=\mathcal{H}om_{f_*\mathcal{O}_X}(f_*\mathcal{M},f_*\mathcal{N})$. In particular, $$ f_*\mathcal{H}om_{\mathcal{O}_X}(f^*\mathcal{N},\mathcal{P}) =\mathcal{H}om_{f_*\mathcal{O}_X}(f_*f^*\mathcal{N},f_*\mathcal{P}). $$ Next, one would need to change from morphisms of $f_*\mathcal{O}_X$-modules to morphisms of $\mathcal{O}_Y$-modules. So let $\eta:\operatorname{id}_{\mathsf{Mod}(\mathcal{O}_Y)}\Rightarrow f_*\circ f^*$ be the unit of the pullback-pushforward adjunction $(f^*\dashv f_*):\mathsf{Mod}(\mathcal{O}_Y)\rightleftarrows\mathsf{Mod}(\mathcal{O}_X)$. Then we have a map

\begin{align*} \tag{2}\label{eq} \mathcal{H}om_{f_*\mathcal{O}_X}(f_*f^*\mathcal{N},f_*\mathcal{P})&\to \mathcal{H}om_{\mathcal{O}_Y}(\mathcal{N},f_*\mathcal{P})\\ \varphi&\mapsto\varphi\circ\eta_\mathcal{N}, \end{align*}

which is well-defined, since it maps $\varphi$ to the composite $$ \mathcal{N}\xrightarrow[\eta_\mathcal{N}]{\mathcal{O}_Y\text{-lineal}} f_*f^*\mathcal{N}\xrightarrow[\varphi]{f_*\mathcal{O}_X\text{-lineal}} f_*\mathcal{P}, $$ and the result is $\mathcal{O}_Y$-lineal since there is a map $f^\flat:\mathcal{O}_Y\to f_*\mathcal{O}_X$. The map \eqref{eq} is a morphism of sheaves since it commutes with restrictions (as $(\varphi\circ\eta_\mathcal{N})|_U=\varphi|_U\circ\eta_\mathcal{N}|_U$). But I don't know if the map \eqref{eq} is bijective. Is this the way of proving \eqref{eq2}? Or there is another way?

Answer 1

Let $V\subset Y$ be an arbitrary open set with preimage $U\subset X$. A section of $f_*\mathcal{H}om_{\mathcal{O}_X}(f^*\mathcal{N},\mathcal{P})$ over $V$ is a section of $\mathcal{H}om_{\mathcal{O}_X}(f^*\mathcal{N},\mathcal{P})$ over $U$, i.e., a map $(f^*\mathcal{N})|_U\to \mathcal{P}|_U$. By adjunction, this is the same thing as a map $\mathcal{N}|_V\to (f_*\mathcal{P})|_V$, which is exactly a section of $\mathcal{H}om_{\mathcal{O}_Y}(\mathcal{N},f_*\mathcal{P})$ over $V$. Everything here is reversible, plays nice with restriction, etc. and therefore gives an isomorphism of sheaves.