Dual of a specific coherent sheaf that is a vector bundle

Let's assume we have two short exact sequence of vector bundles on a smooth 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.

You can show that any vector bundle on $$X\times (\mathbb{A}^2\setminus \{(0,0)\})$$ extends uniquely to a vector bundle on $$X\times \mathbb{A}^2$$. The unique extension is given by the double dual of the pushforward or in our case it is $$M^{\vee \vee}$$. My ultimate goal is to understand $$M^{\vee \vee}|_{x=0}$$ and $$M^{\vee \vee}|_{y=0}$$. I have a guess for these restrictions but I am not able to prove them precisely.

My guess: $$M^{\vee \vee}|_{x=0}\cong W_1\oplus W_2$$ and $$M^{\vee \vee}|_{y=0}=W_3\oplus W_4$$. Where all $$W_i$$ s are deformations of a subbundle to the split exact sequence. The kernels appearing in $$W_1$$ and $$W_3$$ are the same (something related to $$V_1\cap V_2$$) and cokernels appearing in $$W_2$$ and $$W_4$$ are the same (something related to $$V/(V_1+V_2)$$). But I don't know how to approach the problem and actually calculate them or verify my guess.

By your definition there is an exact sequence $$0 \to M^\vee \to V^\vee \to V_1^\vee \otimes \mathcal{O}/x \oplus V_3^\vee \otimes \mathcal{O}/y \to 0\tag{*}$$ on $$X \times \mathbb{A}^2$$ (where $$V$$ and $$V_i$$ denote the pullbacks of the same named bundles). Restricting to $$X \times \{x = 0\}$$ one obtains an exact sequence $$0 \to V_1^\vee \otimes \mathcal{O}/x \to M^\vee \otimes \mathcal{O}/x \to V^\vee \otimes \mathcal{O}/x \to V_1^\vee \otimes \mathcal{O}/x \oplus V_3^\vee \otimes \mathcal{O}/(x,y) \to 0.$$ The morphism $$V^\vee \otimes \mathcal{O}/x \to V_1^\vee \otimes \mathcal{O}/x$$ is the restriction of the morphism dual to $$V_1 \to V$$, therefore it is surjective and its kernel is $$V_2^\vee \otimes \mathcal{O}/x$$. This means that $$M^\vee\vert_{x = 0}$$ is an extension of the kernel of the morphism $$V_2^\vee\vert_{x = 0} \to V_3^\vee\vert_{x = y = 0}$$ (the morphism is induced by the composition $$V_3 \to V \to V_2$$) by the bundle $$V_1^\vee\vert_{x = 0}$$.
EDIT. Let me explain how the sequence $$(*)$$ is obtained.
First, note that there is an exact sequence $$0 \to V_1 \oplus V_3 \stackrel{(x,y)}\to V_1 \oplus V_3 \to V_1 \otimes \mathcal{O}/x \oplus V_3 \otimes \mathcal{O}/y \to 0.$$ Therefore, by definition of pushout there is an exact sequence $$0 \to V \to M \to V_1 \otimes \mathcal{O}/x \oplus V_3 \otimes \mathcal{O}/y \to 0.$$ Dualizing it one obtains $$(*)$$.
• Thanks for you answer. How did you get the first exact sequence? I know dualizing is left exact and turns the pushout into a pullback. I'm not sure how the surjectivity of your short exact sequence works. By $V_3^{\vee}|_{x=y=0}$ do you mean pukkback from $x=y=0$ to $x=0$? because one is defined on $x=0$ and the other one on $x=y=0$. – user127776 Nov 18 '20 at 19:09
• I added an explanation about the first sequence. By $V_3^\vee\vert_{x = y = 0}$ I mean the pullback with respect to the composition $X \times \{x = y = 0\} \to X \times \mathbb{A}^2 \to X$. Note that this map is an isomorphism, so this is essentially you original bundle $V_3$. – Sasha Nov 18 '20 at 19:57
• Alright. But then, the two vector bundles $V_2^{\vee}|_{x=0}$ and $V_3^{\vee}|_{x=y=0}$ are not defined on the same space. How is there a map between them? – user127776 Nov 18 '20 at 20:05