The Stacks project

Is the functor $f^{-1}$ on sheaves of rings truly exact, or simply right-exact (as is the case for abelian sheaves)? This doesn't change the content of the proof, however.

What was meant was that $f^{-1}$ is an exact functor on the category of sheaves. But I also do not understand your comment as $f^{-1}$ is an exact functor on the category of abelian sheaves.

In case anyone needs it, here's a very detailed proof of the last three equalities:

Along the proof we will use the adjunction $$\tag{1}\label{1} \operatorname{Hom}_{\mathcal{O}}(\mathcal{O}[\mathcal{F}],\mathcal{G})\cong\operatorname{Hom}(\mathcal{F},\mathcal{G}),$$ pointed out in the second paragraph of this comment.

Lemma 1. Let $\mathcal{O}$, $\mathcal{F}$ be sheaves over $X$, respectively, of rings and of sets, and let $x\in X$ be a point. There is an isomorphism of $\mathcal{O}_x$-modules $\mathcal{O}[\mathcal{F}]_x\cong\mathcal{O}_x[\mathcal{F}_x]$, that is natural on $\mathcal{F}$.

Proof. Consider the unit of the adjunction $\mathcal{F}\to\mathcal{O}[\mathcal{F}]$ and take map on stalks $\mathcal{F}_x\to\mathcal{O}[\mathcal{F}]_x$. This induces an $\mathcal{O}_x$-linear map $\mathcal{O}_x[\mathcal{F}_x]\to\mathcal{O}[\mathcal{F}]_x$ (natural on $\mathcal{F}$, by naturality of the unit). Namely, it maps $[s_x]$ to $[s]_x$. Conversely, for each open neighborhood $U\subset X$ of $x$, there is an obvious map $\mathcal{F}(U)\to\mathcal{O}_x[\mathcal{F}_x]$. Since $\mathcal{O}_x[\mathcal{F}_x]$ is an $\mathcal{O}(U)$-module, this induces a map $\mathcal{O}(U)[\mathcal{F}(U)]\to\mathcal{O}_x[\mathcal{F}_x]$. Since the latter is compatible with restrictions to open neighborhoods of $x$, we obtain a canonical map $\mathcal{O}[\mathcal{F}]_x\to\mathcal{O}_x[\mathcal{F}_x]$. But it maps $[s]_x$ to $[s_x]$, so the two maps are mutually inverse. $\square$

Lemma 2. Same notation as in Lemma 1, and suppose that $\mathcal{G}$ is a sheaf of $\mathcal{O}$-modules. Then the following square commutes: $$\require{AMScd} \begin{CD} \operatorname{Hom}_{\mathcal{O}}(\mathcal{O}[\mathcal{F}],\mathcal{G}) @>{\cong}>>\operatorname{Hom}(\mathcal{F},\mathcal{G})\\ @VVV@VVV\\ \operatorname{Hom}_{\mathcal{O}_x}(\mathcal{O}_x[\mathcal{F}_x],\mathcal{G}_x) @>>{\cong}>\operatorname{Hom}(\mathcal{F}_x,\mathcal{G}_x) \end{CD}$$ where the left arrow is obtained from the induced map on stalks plus the isomorphism of Lemma 1.

We leave the proof to the reader. In the diagram, the horizontal maps are \eqref{1} and the universal property of the free module over a set.

Lemma 3. Notations as in Lemma 1. Let $f:Y\to X$ be a continuous map of topological spaces. Then $f^{-1}(\mathcal{O}[\mathcal{F}])\cong f^{-1}\mathcal{O}[f^{-1}\mathcal{F}]$.

Proof. Pulling back the unit of the adjunction $\eta:\mathcal{F}\to\mathcal{O}[\mathcal{F}]$ gives a map $f^{-1}\mathcal{F}\to f^{-1}(\mathcal{O}[\mathcal{F}])$. Note that $f^{-1}(\mathcal{O}[\mathcal{F}])$ is a $f^{-1}\mathcal{O}$-module. By the adjunction \eqref{1}, we get a map $f^{-1}\mathcal{O}[f^{-1}\mathcal{F}]\to f^{-1}(\mathcal{O}[\mathcal{F}])$. On stalks, this map equals $$\label{2}\tag{2} (f^{-1}\mathcal{O})_x[(f^{-1}\mathcal{F})_x]\to f^{-1}(\mathcal{O}[\mathcal{F}])_x,$$ where we have used Lemma 1. By Lemma 2, \eqref{2} is adjoint to $(f^{-1}\mathcal{F})_x\to f^{-1}(\mathcal{O}[\mathcal{F}])_x$, i.e., to $\eta_x:\mathcal{F}_x\to \mathcal{O}[\mathcal{F}]_x$. In other words, the map \eqref{2} equals the canonical morphism $\mathcal{O}_x[\mathcal{F}_x]\to \mathcal{O}[\mathcal{F}]_x$ of Lemma 1, which is an isomorphism. $\square$

Going to leave as is.