$\newcommand{\OX}{\mathcal{O}_X}\newcommand{\Hom}{\mathcal{H}\text{om}}\newcommand{\C}{\mathcal}\newcommand{\OXM}{\OX\text{-mod}}$ Are the following three things true?
- $\hom_{\OXM}(\OX,\C{F})\cong F(X)$ so that $\hom(\OX,-)=\Gamma(X,-)$ are naturally equivalent functors. (Quite confident of this, since the commuting diagrams encoding the $\OX$-module structure ensure that the image of any section $\OX(U)\to \C{F}(U)$ is determined by where one sends $1_X\in \OX(X)$, i.e. $s_U\in \OX(U)$ maps to $s_U\cdot \text{res}^X_U\varphi_X(1_X)$).
- $\Hom(\OX,\C{F})\cong \C{F}$ i.e. that $\Hom(\OX,-)$ is naturally equivalent to the identity functor $\text{id}:\OXM\to \OXM$. (This should follow immediately from 1, noting that $\OX|_U = \C{O}_U$ 'local global sections')
- Being that $\Hom(\OX,-)$ is the identity functor, it is exact, so although it is weird to take left or right derived functors,it certainly is still well defined. But am I correct that $\Bbb{R}\Hom(\OX,-)(\C{F}^\bullet) = \C{H}^i(\C{F}^\bullet)[-i]$ for $\C{F}^\bullet \in D(\OXM)$? Similarly, surely if we take the identity morphism of schemes $\text{id}:X\to X$ then $\text{id}_*:\OXM\to\OXM$ is the identity pushforward, and $R^i \text{id}_*=\C{H}^i(-)[-i]$?
Related questions: assuming 2 is true, is it normal for the internal hom out of the monoidal unit to be the identity functor? So the derived functor associated to $\hom_{\OXM}(\OX,-)$ gives the abelian group 'global section' cohomology, i.e. we take K-injective resolutions and then apply global sections, and then take cohomology as $\OX(X)$-modules. Vs derived functor of $\Hom_{\OXM}(\OX,-)$ gives cohomology sheaves in each degree. Probably this is wrong since I've never seen anyone write that down.