Let's assume we have two short exact sequence of vector bundles on some nice variety $X$ over a field. $l_1: 0\rightarrow V_1 \rightarrow V \rightarrow V_2 \rightarrow 0$ and $l_2: 0\rightarrow V_3 \rightarrow V \rightarrow V_4 \rightarrow 0$. Let's pullback everything to $X\times \mathbb{A}^2$. Now we have a map from $V_1\oplus V_3$ to $V$ which at each component is the inclusion. Consider the pushout of this map along the map from $V_1\oplus V_3$ to itself that is multiplication by $x$ on first component and $y$ on the second one. Note that $x$ and $y$ correspond to the variables of the $\mathbb{A}^2$. Let's call this pushout $M$. $M$ is a coherent sheaf on $X\times \mathbb{A}^2$ that is a vector bundle on $X\times (\mathbb{A}^2\setminus \{(0,0)\})$. Around $X\times\{(0,0)\}$ things get complicated. Still it is easy to write down $M|_{x=0}$ or $M|_{y=0}$. Each of these restriction is the direct sum of two coherent sheaves. One of them is a vector bundle that is either $V_1$ or $V_3$ pulled back to $X\times \mathbb{A}^1$. The other summand is something that is a little bit similar to the deformation of a submodule of $V_2$ or $V_4$ to a direct sum. But you can write down them exactly. My question is what is the restriction of dual of $M$, i.e. $M^{\vee}$, to either of the lines $x=0$ or $y=0$. You can assume that $M^{\vee}$ is a vector bundle on $X\times \mathbb{A}^2$ if that helps.