I'm trying to solve the following problem:
Let $M$ be a smooth manifold and $\pi: E \to X$ and $\pi ':F \to X$ be smooth vector bundles of $X$. Suppose we have a (smooth) bundle morphism $f:E\to F$ of constant rank. Show that $\pi_{ket(f)}:ker(f) \to X$ is a sub-bundle of $\pi$.
Here $Ker(f)= \cup_{x \in X}ker(f_{x})$ where $f_{x}=f|_{E_{x}}$ the restriction of $f$ to the fibre over $x$.
I'm not sure how to show two things:
(i) That $ker(f)$ is closed in $X$. (I'm using a definition in which sub-manifolds are closed).
(ii) Suppose I have local trivializations $\{(U,h)\}$ of $\pi: E \to X$. Let the rank $E$ be $r$ and suppose that $dim(ker(f_{x}))=s$ for all $x$. We have $$ h|_{ker(f)\cap\pi^{-1}(U)}:ker(f) \cap \pi^{-1}(U)\to U \times \mathbb{R}^{s}. $$ I know that I have $$ h:\pi^{-1}(U)\to U \times \mathbb{R}^{r}. $$ and that for each $x$ we have $E_{x}\cap ker(f_{x})$ is a vector subspace of dimension $s$ so that if we identify $E_{x}$ with $\mathbb{R}^{r}$ we can choose a basis of $E_{x}$ so that $E_{x}\cap ker(f_{x})$ can be identified with $R^{s}$. I'm just not sure how to find a smooth choice of change of basis that varies with $x$.
Any help would be greatly appreciated. Thank you!