Suppose we have vector bundles $\pi_E:E\to M$ and $\pi_{E'}:E'\to M'$, smooth submersion $G:M\to M'$ and vector bundle morphism $F:E\to E'$ over $G$. Are then the sets $ker F$ and $im F$ smooth vector subbundles of $E$ and $E'$.
I know that the answer is yes if $G=id_M$, however I am not sure what the answer is in this case. Can someone give me some literature with the proof please.
EDIT: I forgot to say that $F$ has a constant rang.
For the case of $\operatorname{ker}(F)$, we can use the fact that $F$ factors uniquely through the pullback bundle $G^*E'$, i.e. there is a unique morphism $\widetilde{F}:E\to G^*E'$ over $\operatorname{id}_M$ such that the $H\circ\widetilde{F}=F$ where $H:G^*E'\to E'$ is the pullback map. Since $H$ is an isomorphism on fibers, $\widetilde F$ is constant rank iff $F$ is.
The $\operatorname{im}(F)$ case is not true without further assumptions. The image of such a vector bundle morphism may fail to restrict to a subspace on a single fiber. For instance, a morphism over the map from the 2-point discrete space to the singleton is a union of two subspaces.