Can we enhance the extension by zero-pullback adjunction to hom sheaves?

Question

$ \def\Mod{\operatorname{Mod}} \def\O{\mathcal{O}} \def\Homs{\mathcal{H}om} \def\Hom{\operatorname{Hom}} \def\F{\mathcal{F}} \def\G{\mathcal{G}} $ Let $X$ be a ringed space, $U\subset X$ be an open subset and $j:U\to X$ be the inclusion. Then the functor $j^{-1}:\Mod(\O_X)\to\Mod(\O_U)$ has a left adjoint, $j_!:\Mod(\O_U)\to\Mod(\O_X)$, the extension by zero (for details, see 009Z for example).

Given a morphism of ringed spaces $f:X\to Y$, the adjunction $f^*:\Mod(\O_Y)\rightleftarrows\Mod(\O_X):f_*$ can be enhanced to a hom sheaf isomorphism $$ f_*\mathcal{H}om_{\O_X}(f^*\mathcal{F},\mathcal{G}) \cong \mathcal{H}om_{\O_Y}(\mathcal{F},f_*\mathcal{G}). $$ (See this post, for example).

I was wondering if something similar could be performed with the adjunction $j_!\dashv j^{-1}$. Maybe something like this? $$ \tag{1}\label{sheaf_adj} \Homs_{\O_X}(j_!\F,\G)\cong j_*\Homs_{\O_U}(\F,j^{-1}\G). $$

Answer 1

$ \def\Mod{\operatorname{Mod}} \def\O{\mathcal{O}} \def\Homs{\mathcal{H}om} \def\Hom{\operatorname{Hom}} \def\F{\mathcal{F}} \def\G{\mathcal{G}} $ The answer is yes: the relation $(1)$ is true. By Corollary 3 of this answer, we have $$ \Homs_{\O_X}(j_!\F,\G)\cong\Homs_{\O_X}(j_{p!}\F,\G), $$ where $j_{p!}\F$ denotes the extension by zero presheaf. We shall prove the latter is isomorphic to $j_*\Homs_{\O_U}(\F,\underbrace{j^{-1}\G}_{\G|_U})$. We are going to change notation: instead of $j$ we will write $j^X_U$, and instead of $j_{p!}$ we will write $j_{U,p!}^X$. For an open set $V\subset X$, we have \begin{align} \Homs_{\O_X}(j_{U,p!}^X\F,\G)(V) &=\Hom_{\O_V}(j_{U,p!}^X\F|_{V},\G|_{V})\\ &=\Hom_{\O_V}(j_{U\cap V,p!}^V(\F|_{U\cap V}),\G|_V)\\ &\cong\Hom_{\O_{U\cap V}}(\F|_{U\cap V},\G|_{U\cap V})\\ &=\Homs_{\mathcal{O}_U}(\F,\G|_U)(U\cap V)\\ &=j_{U*}^X\Homs_{\mathcal{O}_U}(\F,\G|_U)(V). \end{align} So we have bijections on sections. Why do they commute with restrictions? This is due to the way the adjunction $j_!\dashv j^{-1}$ works. See 00A3, for example.