Let $f : M\to N$ be a submersion with $M$ compact and $N$ connected. The $f$ is surjective.

Question

I have no idea how to do this. I tried think in, once $N$ is connected and locally path connected it has to be path connected, but, this does not help. Any hints, solutions, will be very appreciate...

Now I realized that $f(M)$ will be also compact... But, how to conclude the claim?

Thanks

score 3 · Accepted Answer · answered Apr 25 '16 at 13:49

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Hints: The image set $f(M)$ will be compact, so closed. And the rank of $f$ will be $= \dim N$ for any $p\in M$, so $f(M)$ will be...

answered Apr 25 '16 at 13:49

Martín-Blas Pérez Pinilla

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I am inclined to say that $f(M) = N$, but, where do I using connectedness? – L.F. Cavenaghi Apr 25 '16 at 13:50
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If $N$ is connected then all the tangent spaces have the same dimension as $N$. – Ailbe Apr 25 '16 at 13:51
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@user Hint 2: you need it to go from the hint to your conclusion that $f(M)=N$. Think about embedding a sphere into the disjoint union of two spheres. – Apr 25 '16 at 13:55
@Sam now I see, thanks a lot. I appreciate your answer. – L.F. Cavenaghi Apr 25 '16 at 14:01
@user43687, thank you! Nice example, I do appreciate! – L.F. Cavenaghi Apr 25 '16 at 14:02

Let $f : M\to N$ be a submersion with $M$ compact and $N$ connected. The $f$ is surjective.

1 Answers1