Can non-split extension be isomorphic to the split one as objects Is it possible to have a non-split short exact sequences of vector bundles (on some smooth variety) $0\rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$. Such that $V_2\cong V_1\oplus V_3$ as vector bundles?
 A: $\newcommand{\cO}{\mathcal{O}}$Consider exact sequence of trivial vector bundles $$0\to\cO\xrightarrow{\left(\begin{matrix}x \\ y\end{matrix}\right)}\cO\oplus\cO\xrightarrow{\left(\begin{matrix}y & -x\end{matrix}\right)}\cO\to 0$$ on $X=\mathbb{A}^2_{x,y}\setminus\{0\}$. One checks easily that it is exact on stalks (but the same sequence on $\mathbb{A}^2$ is not exact at $(x,y)=(0,0)$). It is the pullback of $0\to\cO(-1)\to \cO\oplus\cO\to \cO(1)\to 0$ from $\mathbb{P}^1$. A splitting of this sequence would be a pair of functions $f_1,f_2\in H^0(\mathbb{A}^2\setminus\{0\},\cO)=k[x,y]$ such that $yf_1-xf_2=1$ but there is no such pair.

On the positive side, any such sequence has to be split if $X$ is proper. In this case the spaces $Hom(V_3, V_1)$ and $Hom(V_3, V_3)$ are finite-dimensional over $k$. Applying $Hom(V_3,-)$ to this exact sequence we get a left exact sequence of finite-dimensional vector spaces $$0\to Hom(V_3, V_1)\to Hom(V_3,V_2)\to Hom(V_3,V_3)$$ The dimension of the vector space in the middle $Hom(V_3,V_2)\simeq Hom(V_3, V_1\oplus V_3)$ is equal to the sum of the dimensions of first and third terms. Therefore, the sequence has to be exact on the right and, in particular, the identity $Id_{V_3}\in Hom(V_3, V_3)$ lifts to a morphism $Hom(V_3, V_2)$ that gives a splitting.
