Kernel of a bundle map Hello!
Let $E$ and $F$ be two vector bundles and let $f:E\rightarrow F$ be a bundle map. Then the kernel of $f$ is not always a subbundle of $E$. Does somebody have a simple example? Does there exist any simple conditions for which the kernel is a subbundle?
Thank you!
 A: Another example. Let $$E = [0,1] \times \mathbb{R} $$ be the trivial line bundle over the unit interval. We have a bundle homomorphism $\phi: E \to E  $ defined by $\phi(x,t)=(x,xt)$ which has kernel $$ \left( \{0\} \times \mathbb{R} \right) \cup \left( (0,1] \times {0} \right).$$
This is not a sub-bundle because it does not have locally-constant rank. 
A: I think this may be an example:
Consider the map from $S^1 \times \mathbb{R}^2 \rightarrow S^1 \times \mathbb{R}^2$ given by the linear map $f_\theta: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that takes the basis vectors $e_1, e_2$ to $e_1, e_2^\theta$, where $e_2^\theta$ is $e_2$ rotated by $\theta$. This should be a map of bundles over $S^1$, but the kernel isn't a subbundle as it has no trivializing neighborhood around, say, $\theta =\pi/2$. 
A: Here is a standard example (which you can find also in the big book of Demailly) in the complex category with the language of sheaves which perhaps may be of some interest for you.
Take $X=\mathbb C^3$, $\mathcal F=\mathcal O^{\oplus 3}$, $\mathcal G=\mathcal O$ and
$$
\varphi\colon\mathcal O^{\oplus 3}\to\mathcal O,\quad (u_1,u_2,u_3)\mapsto\sum_{j=1}^3z_j\,u_j(z_1,z_2,z_3).
$$
Since $\varphi$ yelds a surjective bundle morphism on $\mathbb C^3\setminus\{0\}$, it is easy to see taht $\ker\varphi$ is locally free of rank $2$ over $\mathbb C^3\setminus\{0\}$. However, by looking at the Taylor expansion of the $u_j$'s at $0$, you can check that $\ker\varphi$ is the $\mathcal O$-submodule of $\mathcal O^{\oplus 3}$ generated by the three sections $(-z_2,z_1,0)$, $(-z_3,0,z_1)$ and $(0,z_3,-z_2)$, and that any two of these three sections cannot generate the $0$-stalk $(\ker\varphi)_0$. 
Thus, $\ker\varphi$ is not locally free.
On the other hand, if you have a surjective bundle morphism, the kernel is always locally free.
