Image of smooth vector bundle morphism Let $\pi_1:V_1 \rightarrow B_1$ and $\pi_2:V_2 \rightarrow B_2$ be smooth vector bundles and write $\phi_V: V_1 \rightarrow V_2$ as well as $\phi_B:B_1 \rightarrow B_2$ respectively for the total and base part of a smooth vector bundle morphism.
The Question is, what additional properties must we assume on $\phi_V$ / $\phi_B$ such that:
1.) The image is a smooth vector bundle?
2.) The image is a smooth sub(vector)bundle?
3.) The preimage $\phi_V^{-1}(W) \rightarrow \phi_B^{-1}(A)$ of a smooth sub(vector)bundle $W \subset V_2 \rightarrow A \subset B_2$ is a smooth sub(vector)bundle of $\pi_1$?
 A: As Zhen has noted then we need $\phi_B$ to be surjective for the image of $V_1$ to have any chance of being a vector bundle. Once this is satisfied then I think it is necessary and sufficient that the rank of $Im(V_1)_x$, as a subspace of $V_{2,\phi_B(x)}$ be the same for all $x$ in $B_1$.
The problem is that it is very hard to give any kind of non-trivial verifiable general condition on the map $\phi_V$ that is equivalent to the previous condition, reducing it to the usefulness of a tautology. Even in the category of analytic manifolds and maps, which is highly rigid, there is no such condition.
An example may help. Let's take as our base $B$ the complex plane $\mathbb C$ and let's consider the trivial vector bundle $V$ with fiber $\mathbb C^n$ over $B$. Then $V$ is just the product $\mathbb C^n \times B$ and $\pi$ is the projection onto the second factor.
Now consider the following endomorphism of $V$. We define $\phi : V \to V$ by setting $\phi(v,z) = (zv, z)$. At a point $z \not= 0$ the image of $\phi$ is all of $\mathbb C^n$. At the point $z = 0$ however, the image of $\phi$ is the trivial space $\{ 0 \}$. We see that the image is not of constant rank and thus not a vector bundle. In fact it isn't even a coherent sheaf in the analytic category, as the rank of the stalks of such objects only "jumps up" (we would need to "sheafify" the image to get such a coherent image sheaf).
