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So I was trying to solve this problem in the Differential Geometry notes of Rui Loja Fernandes:

Let $\pi: M \to Q$ be a surjective submersion, $\Phi:M \to N$ and $\Psi: Q \to N$ any maps into a smooth manifold $N$ such that the following diagram commutes:

I want to show that $\Phi$ is smooth iff $\Psi$ is smooth. So one direction is easy: if $\Psi$ is smooth, then $\Phi$ is smooth as the composition of smooth functions.

On the other hand, suppose $\Phi$ is smooth, and let $q \in Q$. We want to show that $\Psi$ is smooth at $q$. As $\pi$ is surjective, there is an $m \in M$ such that $q=\pi(m)$. By the local form for submersions, locally $\pi$ can be written as $\pi(x^1,\dots,x^d)=(x^1,\dots,x^e)$, where $d=\dim M, e = \dim Q$. So locally we have $\Phi(x^1,\dots,x^d)=\Psi(x^1,\dots,x^e)$. Is this enough to prove smoothness of $\Psi$?

ImHackingXD
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1 Answers1

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Yes. A perhaps more elegant way to phrase it is: smooth surjective submersions admit smooth local sections around each point in the codomain (proof: local form for submersions). Here, "section" means "right-inverse".

Using that smoothness is a local concept, the argument goes like this:

  • If $\Psi$ is smooth, then $\Phi = \Psi\circ \pi$ is a composition of smooth functions, hence smooth.
  • If $\Phi$ is smooth, let $x\in Q$ and fix an open neighborhood $U\subseteq Q$ of $x$ together with a local section $s\colon U\to M$; then $\Psi|_U = \Phi\circ s$ is a composition of smooth functions, hence smooth. As $x$ was arbitrary, $\Psi$ is smooth.

Again, the above argument is not really different than the one you gave; you can think that in coordinates one has $s(x^1,\ldots, x^e) = (x^1,\ldots, x^e,0,\ldots, 0)$, according to your notation. But I feel it is better to "box" the existence of local sections as a lemma, and not explicitly use coordinates later.

Ivo Terek
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