# 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 $\pi_1$?

• What's the difference between (1) and (2)? What's your definition of a morphism of smooth vector bundles? Dec 27, 2011 at 2:20
• A smooth vector bundle morphism is a pair of smooth maps that respects the linear and the bundle structure. (Look for example at Wikipeadia or are there others?) Are 1.) and 2.) different? Thats part of the question. If not why can't we assume a priory that the smooth structures are equal? Dec 27, 2011 at 2:58
• Well, that's precisely the point: (1) is not interesting. We want the smooth structure the image to be induced from the smooth structure of $V_2$ – so we really only care about (2). Dec 27, 2011 at 3:16
• Ok. I agree with that because otherwise we get problems with the smoothness of the maps. So question 1.) is equivalent to question 2.) Dec 27, 2011 at 3:19

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).