I am confused about the following. I know that for two line bundles $L\_1, L\_2$ on an algebraic curve $C$ the vector space ${\rm Ext}^1(L\_1,L\_2)$ classifies isomorphism classes of rank two vector bundles on $C$ which are extensions of $L\_2$ by $L\_1$. My question is, does this mean that there is a "universal" rank two vector bundle $E$ on ${\rm Ext}^1(L\_1,L\_2) \times C$ ? 


I.e. if I let $p\_2 : {\rm Ext}^1(L\_1,L\_2) \times C \to C$ be the projection then does there exist a rank two vector bundle $E$ on ${\rm Ext}^1(L\_1,L\_2) \times C$ such that $0 \to p\_2^\*L\_1 \to E \to p\_2^\*L\_2 \to 0 $ is a short exact sequence and for all $x \in {\rm Ext}^1(L\_1,L\_2)$ if we restrict the above short exact sequence to $x \times C$ we get the exact sequence $0 \to L\_1 \to E\_x \to L\_2 \to 0 $ of vector bundles on $C$ which corresponds to the extension class $x$ ? 



${\rm Ext}^1$ and extensions of line bundles on a curve
I wanted to work this out for myself anyways, so here's a summary of the argument in Le Potier.


Suppose $X$ is a projective variety, and $E,G$ are locally free sheaves on $X$. Put $S = {\rm{Ext}}^1(G,E)$, and let ${\bf E},{\bf G}$ be the constant families on $S\times X$, namely ${\bf E} = p\_2^\ast E$ and ${\bf G} = p\_2^\ast G$. We want to construct a universal extension


$$0\to {\bf E} \to {\bf F} \to {\bf G}\to 0.$$


We can specify $\bf F$ by giving an element of ${\rm Ext}^1({\bf G},{\bf E})$.


What is ${\rm Ext}^1({\bf G},{\bf E})$? Intuitively, we can specify an extension of ${\bf G}$ by ${\bf E}$ by specifying for each $s\in S$ an extension of $G$ by $E$; that is, it should be the space of morphisms $S\to S$; furthermore, if $f:S\to S$ and ${\bf F}(f)$ is the sheaf corresponding to $f$, then, ${\bf F}(f)\_s$ is the extension of $G$ by $E$ corresponding to the extension class $f(s)$. More formally, considering the diagram


\begin{array}{ccc} S\times X &\stackrel{p\_2}{\to} &X\\ \downarrow {p\_1} & & \downarrow p\_1'\\
S & \stackrel{p\_2'}{\to} & pt\end{array}


we find


\begin{align}{\rm Ext}^1({\bf G},{\bf E})&= H^1( {\mathcal H}om({\bf G},{\bf E})) \\&= H^0(R^1 p\_{1\*} {\mathcal H}om({\bf G},{\bf E})))\\
&= H^0(p\_2'^\* R^1 p'\_{1\*} {\mathcal H}om(G,E))\\
&= H^0(\mathcal O\_S \otimes {\rm Ext}^1(G,E)),\end{align}


making use of various identities from Hartshorne III.6-9. The desired universal family corresponds to the distinguished identity morphism $S\to S$.


