Let $\mathcal{F}$ and $\mathcal{G}$ be vector bundles on a smooth projective scheme $X$. Let $\varphi : \mathcal{F} \longrightarrow \mathcal{G}$ be a morphism of vector bundles with ranks $m$ and $n$, respectively, and $m > n$.
The set $$D_{k}(\varphi) = \bigl \{ x \in X ; \text{rank}(\varphi(x))\leq k \bigr \}$$ where $k \leq \text{min}\bigl \{ m, n \bigr \}$ is called degeneracy locus of rank $k$ of $\varphi$.
Suppose that $\varphi_{x} : \mathcal{F}_{x} \longrightarrow \mathcal{G}_{x}$ is not surjective for all $x \in X$. Let $U$ be a dense open subset of $X$.
To show that $\varphi$ is generically surjective, we must show that $\varphi_{p} : \mathcal{F}_{p} \longrightarrow \mathcal{G}_{p}$ is surjective for all $p \in U$. Right or wrong? Does information about the $D_{k}(\varphi)$ set allow us to show this? What is the best strategy to show that such morphism is generically surjective?
Thanks in advance for the suggestions.
Let $V\subset X$ be a small open subset where both $\mathcal{F}$ and $\mathcal{G}$ are trivial. After choosing trivializations $\varphi|_V \in {\rm Hom}(\mathcal{F}, \mathcal{G})|_V$ is given by a $n \times m$ matrix $A_V$ with entries in $\mathcal{O}_X(V)$. Hence for every $x\in V$, $$ {\rm rk}\ \varphi_x = {\rm rk}\ A_V(x) $$ In particular, it can be computed using the minors of $A_V(x)$. Indeed $D_{k-1}(\varphi) \cap V$ (in your notation) is given by the vanishing of the $k\times k$ minors of $A_V$.
We now arrive at the following equivalence: $\varphi$ is generically surjective if and only if $D_{n-1}(\varphi) \neq X$. Indeed $\varphi_x$ is surjective if and only if ${\rm rk}\ \varphi_x = n$ if and only if $x \in X \setminus D_{n-1}(\varphi)$. On the other hand $D_{n-1}(\varphi)$ is a closed subvariety of $X$.