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?