If there is a short exact sequence of holomorphic vector bundles, $$0 \overset{a_1}{\to} W \overset{a_2}{\to} V \overset{a_3}{\to} F \overset{a_4}{\to} 0,$$ then one can expect a $C^{\infty}$ splitting $$V \cong W \oplus F$$ rather than a holomorphic splitting.
I know that a s.e.s. needs consecutive maps to equal $1$, and that for exactness that $im(a_i) = ker(a_{i+1})$. I also know that a vector bundle is just a manifold with the fiber as a vector space (complex here). For a shorthand of notation of a vector bundle, I use $\pi: E \to B$ where $B \times V$ is the product space and $\pi$ is the fiber bundle. Written like a s.e.s., this is $$V \to E \overset{\pi}{\to} B.$$ Also $a_2$ is injective and $a_3$ is surjective.
So is the reason why the splitting is only $C^{\infty}$, and not holomorphic, because the maps, either $a_2^{-1}$ or $a_3^{-1}$ are not injective?
Best Answer
On a (paracompact) complex manifold all short exact sequences of $C^{\infty}$ vector bundles are $C^{\infty}$ split, so it is enough to exhibit an exact sequence of holomorphic vector bundles that doesn't holomorphically split.
The simplest example is the exact sequence on $\mathbb P_\mathbb C^1$:
$$0\to \mathcal O(-2) \to \mathcal O(-1) \oplus \mathcal O(-1)\to \mathcal O\to 0$$ It does not split because the bundles $\mathcal O(-1) \oplus \mathcal O(-1)$ and $\mathcal O(-2) \oplus \mathcal O$ are not isomorphic: the second has nonzero holomorphic sections but the first doesn't.