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Let $f:E\rightarrow F$ be a surjective morphism of holomorphic vector bundles on a complex manifold $X$. How I can prove that for each holomorphic section $s: U\rightarrow F$ there exists a section $t:V\rightarrow E$ ($V\subseteq U$) such that $f\circ t=s_{|V}$ ?

Any help is very apreciated.

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    I believe you're supposed to invert the transition function. In any coordinate chart, the transition function is a surjective linear map at each fiber. Some $n\times n$ submatrix is invertible at $p$ and in a nbhd of $p$ by continuity. Invert that submatrix. – 2017-01-09
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    Can you expande your answer? – 2017-01-09

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For any neighborhood $U$ of $p\in X$ over which there are local trivializations for both bundles, in the corresponding coordinate charts the map $f$ has representation

$$U\times\mathbb{C}^n\to U\times\mathbb{C}^m$$

$$(u,v)\to(u,\tau(u)\cdot v)$$

where $\tau$ takes values in the space of $m\times n$ matrices.

Since $f$ is surjective and a bundle homomorphism it is surjective when restricted to each fiber. So for any $u_0$, $\tau(u_0)$ is an $m\times n$ matrix that represents a surjective linear map. Thus the matrix $\tau(u_0)$ is of maximal rank and must have an invertible $m\times m$ submatrix.

Assume $\tau(u_0)$ is of the form $(A\ B)$ with $A$ invertible. By continuity, that same submatrix is invertible for all $\tau(u)$ in a possibly smaller neighborhood $V$ of $u_0$. So the coordinate representation is now $(u,v)\to(u,(A(u)\ B(u))\cdot v)$ where $A$ and $B$ are matrices and $A$ is invertible. We can invert that submatrix $A$ for all $u\in V$ to get a holomorphic function $\phi:V\to GL(m,\mathbb{C})$ sending $u\to A\to A^{-1}$. Ie. $\phi(u)=A(u)^{-1}$.

Let the section $s$ have coordinate representation $(u,r(u)):U\to U\times\mathbb{C}^m$. Now consider the map

$$t:V\to V\times\mathbb{C}^m\to V\times\mathbb{C}^n$$

given by

$$\Bigg(Id\times\binom{\phi(u)}{0}\Bigg) \circ s$$

sending

$$u\to(u,r(u))\to\Bigg(u,\binom{\phi(u)\cdot r(u)}{0}\Bigg)$$

The map $t$ is a local section of $E$. Composing with $f$

$$(f\circ t)(u)=f\Bigg(u,\binom{\phi(u)\cdot r(u)}{0}\Bigg) = \Bigg(u,\tau(u)\cdot\binom{\phi(u)\cdot r(u)}{0}\Bigg)$$

$$= \Bigg(u,(A(u)\ B(u))\cdot\binom{A(u)^{-1}\cdot r(u)}{0}\Bigg)$$

$$=(u, A(u)\cdot A(u)^{-1}\cdot r(u) + B(u)\cdot 0)$$

$$(u,r(u)) = s(u)$$