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Let $(X,\mathcal{O}_X)$ be a ringed space, $\mathcal{F}$ and $\mathcal{G}$$\,\,\,$$\mathcal{O}_X$-modules, and $\varphi:\mathcal{F}\to\mathcal{G}$$\,$ an $\mathcal{O}_X$-module homomorphism.

If $\varphi$ is both injective and surjective, then how do we go about constructing an inverse $\varphi^{-1}:\mathcal{G}\to\mathcal{F}\,$?


For $U\subseteq X$ open, we have a map $\varphi(U):\mathcal{F}(U)\to\mathcal{G}(U)$ and we set$$\begin{align*}\text{Im}(\varphi)(U)=\{s\in\mathcal{G}(U):&\forall x\in U,\exists V\subseteq U\text{ open such that}\\&x\in V,s\hspace{-0.13cm}\mid_V\in\text{Im}(\varphi(V))\}\end{align*}$$ Since $\varphi$ is surjective we have $\text{Im}(\varphi)(U)=\mathcal{G}(U)$, so given $s\in\mathcal{G}(U)$ we can find $U=\cup_{i\in I}V_i$ for some open $V_i\subseteq U$, with $t_i\in\mathcal{F}(V_i)$ such that $\varphi(V_i)(t_i)=s\mid_{V_i}$.

I’d like to patch the $t_i$ together to get some $t\in\mathcal{F}(U)$, so we can then set $\varphi(U)^{-1}(s)=t$. I imagine this is where we need to use the injectivity of $\varphi$, but I can’t seem to spot how.

Any help would be much appreciated.

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  • $\begingroup$ Apologies, I've only just met $\mathcal{O}_X$-modules and I'm not familiar with what that means in this context, have I somewhere assumed that for sufficiently small $U$ we have $\mathcal{F}(U)=\oplus\mathcal{O}_X(U)$? $\endgroup$ – Dave May 16 at 11:30
  • $\begingroup$ I'm just saying most sheaf of $O_X(U)$-modules you want to consider in scheme theory are locally free in which case the injective+surjective implies invertible is easy $\endgroup$ – reuns May 16 at 11:32
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Given $t_i,t_j$, we would have that $\varphi(V_i\cap V_j)(t_i\hspace{-0.13cm}\mid_{V_i\cap V_j})=\varphi(V_i\cap V_j)(t_j\hspace{-0.13cm}\mid_{V_i\cap V_j})=s\mid_{V_i\cap V_j}$, and so by the injectivity of $\varphi$ we have $t_i\hspace{-0.13cm}\mid_{V_i\cap V_j}=t_j\mid_{V_i\cap V_j}$ for all $i,j\in I$, and so we can glue the $t_i$ to construct $t$.

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