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?


1 Answer 1


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 $Z$.

  • $\begingroup$ Why is $w'$ the image of some $w\in TS$? $\endgroup$
    – Tom
    Commented Mar 25, 2015 at 15:24
  • $\begingroup$ 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 $M\times N\cap S'$. $\endgroup$ Commented Mar 25, 2015 at 15:46
  • $\begingroup$ 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. $\endgroup$
    – Tom
    Commented Mar 25, 2015 at 15:57
  • $\begingroup$ You're sure right! Sorry for mixing up! $\endgroup$
    – Jesus RS
    Commented Mar 25, 2015 at 16:07
  • $\begingroup$ @ArturAraujo Do you know a reference for a proof that pullbacks of submersions are submersions or could you even provide one? $\endgroup$
    – Tom
    Commented Mar 25, 2015 at 16:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .