Neutral element in $\operatorname{Ext}(\mathcal{F},\mathcal{G})$ Let $(X, \mathcal{O}_X)$ a ringed space and $ \mathcal{F}, \mathcal{G} \in (\mathcal{O}_X\mathcal{-Mod})$ two arbitrary $\mathcal{O}_X$-modules. The canonical bifunctors
$$\operatorname{Ext}^i(-,-): (\mathcal{O}_X\mathcal{-Mod})^{op} \times \mathcal{O}_X\mathcal{-Mod} \to \operatorname{Ab}$$
are derived functors of the bifunctor $\operatorname{Hom}_{\mathcal{O}_X}(-,-): (\mathcal{O}_X\mathcal{-Mod})^{op} \times \mathcal{O}_X\mathcal{-Mod} \to \operatorname{Ab}$. 
On can show that $\operatorname{Ext}^1(\mathcal{F},\mathcal{G})= \operatorname{Ext}(\mathcal{F},\mathcal{G})$ has a interpretation as set of classes of extensions $0 \to \mathcal{F} \to ? \to \mathcal{G} \to 0$. But it is moreover claimed that $\operatorname{Ext}(\mathcal{F},\mathcal{G})$ has structure of abelian groups. 
How the group operation in is defined and what is neutral element in $\operatorname{Ext}(\mathcal{F},\mathcal{G})$? As a natural guess for the neutral element can be considered the splitting extension $0 \to \mathcal{F} \to \mathcal{F} \oplus \mathcal{G} \to \mathcal{G} \to 0$ since this extension always exist und therefore it can be considered as a kind of 'canonical' choice. That's of course only a conjecture. I'm seeking for a formal argument why this choice indeed gives the neural element of $\operatorname{Ext}(\mathcal{F},\mathcal{G})$.
 A: Let $\mathcal{E,E'}$ be two extensions of $\mathcal{G}$ by $\mathcal{F}$ :
$$0\to \mathcal{F}\xrightarrow{f}\mathcal{E}\xrightarrow{g}\mathcal{G}\to 0$$
$$0\to \mathcal{F}\xrightarrow{f'}\mathcal{E'}\xrightarrow{g'}\mathcal{G'}\to 0$$
Consider the maps $i:\mathcal{F}\to\mathcal{E\oplus E'}$ given on a section $s$ by $i(s)=(f(s),-f'(s))$, and the map $p:\mathcal{E\oplus E'}\to\mathcal{G}$ given by $p(s,s')=g(s)-g'(s')$. The Baer sum of $\mathcal{E,E'}$ is then given by $\ker p/\operatorname{im} i$.
Now assume $\mathcal{E'}$ is the trivial extension :
$$0\to\mathcal{F}\xrightarrow{j}\mathcal{F\oplus G}\xrightarrow{q}\mathcal{G}\to 0$$
Then $\ker p$ is the subsheaf of $\mathcal{E\oplus F\oplus G}$ of sections $(s_1,s_2,s_3)$ such that $g(s_1)-q(s_2,s_3)=0$ or equivalently $g(s_1)=s_3$.
Define a map $u:\ker p\to \mathcal{E}$ such that $(s_1,s_2,s_3)\mapsto s_1+f(s_2)$. We have $u\circ i(s)=u(f(s),-s,0)=f(s)-f(s)=0$ so the map $u$ factors as a map $\ker p/\operatorname{im}i$.
Check that $u$ is a morphism of extensions. As such, it is automatically an isomorphism. But for sake of completeness, we can write its inverse : it is given by $v:\mathcal{E}\to \mathcal{E\oplus F\oplus G}/\operatorname{im}i$ given by $v(s)=(s,0,g(s))$. Check that $p\circ v=0$ so that $v$ factor through $\ker p/\operatorname{im}i$.
We trivially have $u\circ v=\operatorname{id}$. We also have $v\circ u(s_1,s_2,s_3)=v(s_1+f(s_2))=(s_1+f(s_2),0,g(s_1+f(s_2)))$. Now $g(s_1+f(s_2))=g(s_1)$ since $g\circ f=0$ and $g(s_1)=s_3$ since $(s_1,s_2,s_3)\in\ker p$. Hence $v\circ u(s_1,s_2,s_3)=(s_1+f(s_2),0,s_3)=(s_1,s_2,s_3)+(f(s_2),-f(s_2),0)$. But $(f(s_2),-f(s_2),0)=i(s_2,s_3)=0$ in $\ker p/\operatorname{im} i$. It follows that $v\circ u=\operatorname{id}$ as expected.
