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Let $M$ be a smooth manifold (without boundary) and $E$ and $E'$ be smooth vector bundles of over $M$.

Let $F:E\to E'$ be a bundle homomorphism.

For each $p\in M$, we define the rank of $F$ at $p$ as the rank of the linear map $F|E_p$, where $E_p$ denotes the fibre over $p$ in $E$.

We say that $F$ is of constant rank if the rank of $F$ is invariant with respect to $p$.

The kernel of a bundle homomorphism $F:E\to E'$ is defined as $\ker F=\bigsqcup_{p\in M}\ker F|E_p$.

THEROEM. The kernel of a constant rank bundle homomorpsim $F:E\to E'$ is a subbundle of $E$.

This theorem can be found in Lee's Introduction to Smooth Manifolds, II Edition, pg 266.

The proof given in Lee's book is via the concept of local frames.

I was wondering whether we can somehow apply the following fact: Let $M$ and $N$ be smooth manifolds and $F:M\to N$ be a constant rank smooth map. Let $q$ be any point in the image of $F$. Then $F^{-1}(q)$ is an embedded submanifold of $M$.

(Thanks to User 43687 for pointing out a mistake in the above.)

That would lead to a shorter proof.

Can somebody help?

Thanks.

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  • $\begingroup$ What do you mean by the kernel of a smooth map. If you mean it's derivative, then this would be a special case of the claim where $E$ is the tangent bundle. $\endgroup$ – user113529 Feb 24 '15 at 20:06
  • $\begingroup$ @user43687 I am sorry. I made a blunder. Let me correct it. $\endgroup$ – caffeinemachine Feb 24 '15 at 20:09
  • $\begingroup$ I don't think you'll be able to use this. Since the total spaces of each bundle is a smooth manifold, certainly $F^{-1}(s(p))$ is an embedded submanifold ($s$ is the zero section, $p$ a point on the base). In fact, it's a subspace of the fiber. The problem is stitching these sub manifolds together smoothly. $\endgroup$ – user113529 Feb 24 '15 at 20:26
  • $\begingroup$ Also, the statement that $F$ has constant rank (as a smooth map between manifolds) means that the derivative $dF$ has constant rank as a map between tangent bundles. Hence, to fit it in to your context, $dF$ would need to have constant rank as a map of the tangent bundles over the TOTAL spaces $E$ and $E^{\prime}$. $\endgroup$ – user113529 Feb 24 '15 at 20:35
  • $\begingroup$ I agree with your remark. I still have a feeling that we can still make it work. But it's just a feeling. $\endgroup$ – caffeinemachine Feb 24 '15 at 20:49
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Perhaps this will help build a bit of intuition. The theorem your citing is about regular points on manifolds and their preimages. The key here is that the theorem says nothing about how these preimages vary as the point moves smoothly on the manifold. With the kernel of a bundle map, ensuring that it is a submanifold involves showing that the sub spaces smoothly vary as we vary the point p. One can see that the only way we can get smoothness is by looking at the smooth structure of the base space of the target bundle. This is exactly what the theorem your citing is missing...smoothness in p

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