I'm trying to prove the equivalence of the following statements:

Suppose $M^m$ and $N^n$ are smooth manifolds, $S\subseteq M\times N$ immersed, and $\pi_M$ and $\pi_N$ the projection maps. TFAE:

1) $S$ is the graph of a smooth function $f\colon M\to N$

2) $\pi_M|_S$ is a diffeomorphism from $S$ onto $M$,

3) For each $p\in M$, the submanifolds $S$ and $\{p\}\times N$ intersect transversely in exactly one point.

I've shown $(1)\iff (2)$, and $(2)\implies (3)$. I can include that work if needed. Right now I'm just trying to show $(3)$ implies either of the other two.

If $S$ and $\{p\}\times N$ intersect transversely in one point, then the inclusion $\iota\colon S\to M\times N$ is transverse to $\{p\}\times N$, and $\iota^{-1}(p\times N)$ is a singleton. Then this means there exists some $(p,q)\in S$ such that $(p,q)\in p\times N$, thus $\pi_M|_S$ is surjective, and it must be injective since $\iota^{-1}(p\times N)$ is a singleton.

The transversality condition implies $$ T_{(p,q)}(p\times N)+d\iota_{(p,q)}(T_{(p,q)}S)\simeq T_{(p,q)}N+T_{(p,q)}(S)=T_{(p,q)}(M\times N) $$

so necessarily $\dim(T_{(p,q)}(S))\geq m$, by dimension considerations.

As a bijection, it's enough to show $\pi_M|_S$ is either of constant rank, or a local diffeomorphism to prove it's a diffeomorphism. As a projection map, I feel like it should be a submersion onto $M$, and that would do the trick.

Note: This is Theorem 6.32 in John Lee's Intro to Smooth Manifolds.

  • $\begingroup$ Yes. Check out the proof of parametrized transversality in Guillemin and Pollack. $\endgroup$ Aug 18, 2014 at 4:03
  • $\begingroup$ @CharlieFrohman Where is that? I'm scanning sections 1.6 and 2.3 on Transverality, but don't see it in either. $\endgroup$
    – Clara
    Aug 18, 2014 at 4:47
  • $\begingroup$ Oops, 1.5 not 1.6 $\endgroup$
    – Clara
    Aug 18, 2014 at 16:35

1 Answer 1


You've got 99% of the proof already. Hint: Suppose $d\pi_M|_S$ is not surjective and think about the isomorphisms $T_{(p,q)} S+ T_{(p,q)}(\{p\} \times N) \cong T_{(p,q)}(M \times N) \cong T_p M \times T_q N$.

Click/rollover for full answer below:

Full answer. Given $(p,q)\in S$ and $v \in T_p M$, transversality implies that $(v,0) \in T_p(M \times N)$ equals $a+b$ for some $a \in T_{(p,q)} S $ and $b \in T_{(p,q)}(\{p\} \times N)$. Since $T_{(p,q)}(\{p\} \times N)$ lies in the kernel of $d\pi_M$, $v$ lies in the image of $d(\pi_M|_S)$: $v=d\pi_M(a+b)=d\pi_M(a)+d\pi_M(b)=d(\pi_M|_S)(a)+0=0.$ So $d(\pi_M|_S)$ is surjective at $(p,q)$.

  • $\begingroup$ Thanks squirrel. How do you know $(v,0)\in T_{(p,q)}S$? Are you writing $(v,0)=a+b$ for $a\in T_{(p,q)}S$, $b\in T_{(p,q)}(\{p\}\times N)$, and then applying $d\pi_N$ to get $0=d\pi_N(a)+d\pi_N(b)$ or something? And then the result follows from $d(\pi_M|_S)(v,0)=v$? $\endgroup$
    – Clara
    Aug 19, 2014 at 4:45
  • $\begingroup$ My phrasing may have been unclear. How's this: $v=d\pi_M(a+b)=(d\pi_M)|_{T_{(p,q)}S} (a)+(d \pi_M)|_{T_{(p,q)}(\{p\}\times N)}(b)=(d\pi_M)|_{T_{(p,q)}S} (a)+0=d(\pi_M|_S)(a).$ $\endgroup$
    – Kyle
    Aug 19, 2014 at 12:43
  • $\begingroup$ Thanks for catching my error. I realized I didn't mean "$(v,0) \in T_{(p,q)}S$" but rather $"(v,\text{[something]}) \in T_{(p,q)}S$". I edited the original post accordingly. $\endgroup$
    – Kyle
    Aug 19, 2014 at 14:17

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