short exact sequence of holomorphic vector bundles splits but not holomorphically, only $C^{\infty}$ 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?
 A: 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.
A: Note that unless $W$ or $F$ is zero, neither $a_2$ nor $a_3$ will be invertible and so there are no such  maps $a_2^{-1}$ or $a_3^{-1}$.
Given such a short exact sequence, there exists a $C^\infty$ splitting essentially because there exist $C^\infty$ partitions of unity: take a cover of $B$ by open sets over each of which you can trivialize $V$, and put a smooth hermitian metric on each of those trivialized bundles. You can then use a partition of unity to patch those metrics together into a metric on the entire bundle $V$. Then you have a hermitian metric on each fibre $V_x$, and so taking the orthogonal complement of $W_x$ you get a splitting $V_x = W_x \oplus W^\perp_x$, which gives a splitting $V = W\oplus W^\perp$ globally since the metric varies smoothly. Finally, we have $W^\perp\cong F$ since both are isomorphic to the quotient bundle $V/W$.
This whole story fails to work in the holomorphic world because in general there are no holomorphic partitions of unity - there just aren't enough holomorphic functions, in contrast to the smooth case.
