Bijective $\mathcal{O}_X$-Module Homomorphisms

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.

• 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 $\mathcal{F}(U)=\oplus\mathcal{O}_X(U)$? – Dave May 16 at 11:30
• 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 – reuns May 16 at 11:32

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 $$t$$.