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Suppose we have smooth manifolds $M,M',N$, a smooth map $f\colon M\rightarrow M'$ and a smooth submanifold $S'\subseteq M'\times N$, such that the projection $\pi_{M'}\colon S'\rightarrow M'$ is a submersion.

A quick calculation shows, that the map $(f\times id)\colon M\times N\rightarrow M\times N'$ is transverse to $S'$, so $S\colon=(f\times id)^{-1}(S)\subseteq M\times N$ is a smooth submanifold.

Is the projection $\pi_M\colon S\rightarrow M$ always a submersion? If not, which are conditions on $f$, such that $\pi_M$ is a submersion?

Tom
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The answer is yes, it is always a submersion. First of all, $\pi:=\pi_M$ is clearly surjective. Also, for any point $s'\in S'$ lying over some $p'\in M'$, the tangent space at $s'$ maps surjectively onto the tangent space at $p'$ (since $\pi':=\pi_{M'}$ is a submersion.) Take then a point $p\in M$ and a vector $v\in T_pM$, and push-it forward to $M$, $v'=f_*v$. From what we've just noted, there must be a point $s'\in S'$ which actually lies in the intersection $f(M)\times N \cap S$, and the tangent space at that point maps surjectively onto the tangent space at $f(p)$. In other words, there is a vector $w'\in T_{s'}S'$ such that $\pi'_*w=v'$, and this $w'$ is the image of some $w \in TS$. Now, it's obvious that this $\pi_*w=v$.

Here's a high-brow proof: $\pi$ is just the pullback of $\pi'$. The pull back of a submersion is a submersion.

EDIT: Here's some details about the above; I'll only sketch the details, since I think that it is a very useful exercise to convince yourself of these facts. First of all, the image of $f\times \mathrm{id}$ clearly is $f(M)\times N$. Also, by definition, $S'=(f\times \mathrm{id})^{-1}(S)$, so what is $f\times \mathrm{id}(S)$? You should convince yourself that it is precisely $f(M)\times N\cap S'$. Then, I claim $(f\times \mathrm{id})_*(TS)=T(f(M)\times N\cap S')$. This should be easy to see, because $f_*f^*T(f(M)\times N\cap S')=T(f(M)\times N\cap S')$.

Now, $s′$ was chosen to be in the intersection $f(M)\times N \cap S′$, and $v′$ is tangent to $M$, so that $w′$ is tangent to $f(M)\times N \cap S'$. Clearly, from the above, $w'$ is then the pushforward of some $w\in TS$. Now, I above incorrectly stated that $\pi_*w=v$. We're a priori not sure of this; what we know is that there is some $w$ that such equality holds (the problem here is $v$ might not be the only preimage of $v'$, so a randomly chosen $w$ might map to some other preimage; on the other hand, again $f\times \mathrm{id}$ pulls back a copy of $w'$ for every preimage of $v'$.)

(I sense this might sound a bit confusing, mostly because I'm being a bit lazy and not writing thing more explicitly. However, if you think of this whole thing as a pullback, the whole thing becomes clear, because I'm basically going through an "element proof" of my highbrow proof.)

Adition: check also:

The tangent space to the preimage of $Z$ is the preimage of the tangent space of $Z$.

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Artur Araujo
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  • Why is $w'$ the image of some $w\in TS$? – Tom Mar 25 '15 at 15:24
  • Because $s'$ was chosen to be in the intersection $M\times N\cap S'$ and $v'$ is tangent to $M$, so that $w'$ is tangent to $M\times N\cap S'$. – Artur Araujo Mar 25 '15 at 15:46
  • So far so good, but why is $w'$ the image of some $w\in TS$? The differential $d(f\times id)\colon TS\rightarrow TS'$ does not need to be surjective. Sorry if I don't get it. Could you maybe elaborate a little bit more? Thank you very much. – Tom Mar 25 '15 at 15:57
  • You're sure right! Sorry for mixing up! – Jesus RS Mar 25 '15 at 16:07
  • @ArturAraujo Do you know a reference for a proof that pullbacks of submersions are submersions or could you even provide one? – Tom Mar 25 '15 at 16:16
  • Tom: are you familiar with category theory, and the proof that the pullback of an epi is an epi? If so, just apply it to the differential. Otherwise, I'll write it down. (I'm pretty sure it is written down somewhere, since this is fundamental for pulling back all sorts of objects). – Artur Araujo Mar 25 '15 at 21:07
  • @ArturAraujo That would of course work, if one knows that the tangent space of the pullback manifold is the pullback of the tangent spaces. Is this obvious? – Tom Mar 25 '15 at 23:37
  • No, we don't know that, and in fact i messed up (sorry, i keep writing these in a hurry.) Of course, the pull back cannot gives the tangent space that f destroys for not being injective. However, it still doesn't matter, because every vector in the intersection pulls back, and $f_* f^*=\mathrm{id}$. This is enough for what you want to prove – Artur Araujo Mar 26 '15 at 12:04
  • I'm writing this on the train, but I promise I'll sit down later to write a decent proof. – Artur Araujo Mar 26 '15 at 12:05
  • The addition clears it all. Thank you. It even proves, that the tangent space of a transversal pullback is the pullback of the tangent spaces. – Tom Mar 27 '15 at 09:57
  • Careful! The preimage is, but it is not true that $f^* f_$ is the identity. Whatever $f_$ kills cannot be brought back to life by the pullback; $f$ has to be transversal immersion. – Artur Araujo Mar 27 '15 at 12:16
  • Also, somehow I kept forgetting to add the surjective to sujective submersion. This qualifier is needed for the highbrow proof. – Artur Araujo Mar 27 '15 at 12:17