I am having a question when learning the holomorphic vector bundles in page 67, Daniel's Complex Geometry An Introduction.
Let $\phi: E\to F$ be a (holomorphic) vector bundle homomorphism. There exist holomorphic vector bundles $\textit{Ker}(\phi)$ and $\textit{Coker}(\phi)$ over X, such that the fibers over $x \in X$ are canonically isomorphic to $\textit{Ker}(\phi): E(x) \to F(x)$ and $\textit{Coker}(\phi): E(x) \to F(x)$, respectively.
But based on my knowledge in differential geometry, the kernel of a bundle map is not necessarily a vector bundle because the dimension of $\textit{Ker}(\phi)$ may vary on $X$. To make the kernel (or cokernel) a vector bundle, we need some conditions like $\phi$ having constant rank.
For example, let $X = \mathbb{C}$, $E = F = \mathbb{C}^2$, $\pi:E=F \to X$ be the projection to the first coordinate, and $\phi: (z,w) \mapsto (z,zw)$ be the bundle homomorphism. Then the kernel of $\phi$ is trivial on $X \setminus \{0\}$ while is $1$-dimensional at the origin.
I am wondering if I misunderstand or miss anything?