Let $M$ and $N$ be smooth manifolds and $\pi:M\rightarrow N$ be a smooth map. A local section of $\pi$ is a a smooth map $\sigma:U\rightarrow M$ defined on some open subset $U\subseteq N$ such that $\pi\circ\sigma=1_U$.

I am reading about this from Lee's Introduction to smooth manifolds. It says the following :

Many of the important properties of smooth submersions follow from the fact that they admit an abundance of smooth local sections.

Local section theorem says the following :

Suppose $M$ and $N$ are smooth manifolds and $\pi:M\rightarrow N$ is a smooth map. Then, $\pi$ is a smooth submersion if and only if every point of $M$ is in the image of a smooth local section of $\pi$.

I was trying to prove this on my own.

Let $p\in M$. As $\pi:M\rightarrow N$ is a smooth submersion, it is in particular of constant rank $n$ (dimension of $M$ is $m$, dimension of $N$ is $n$). So, constant rank theorem says that given this $p$, there exists chart $(U,\varphi)$ cetered at $p$ and a chart $(V,\psi)$ centered at $q=\pi(p)$ such that $$\psi\circ \pi\circ \varphi^{-1}:\varphi(U)\subseteq \mathbb{R}^m\rightarrow \psi(V)\subseteq \mathbb{R}^n$$ is of the form $$(x_1,\cdots,x_n,x_{n+1},\cdots,x_m)\mapsto (x_1,\cdots,x_n).$$

This map $\mathbb{R}^m\rightarrow \mathbb{R}^n$ given by $(x_1,\cdots,x_n,x_{n+1},\cdots,x_m)\mapsto (x_1,\cdots,x_n)$ have so many global sections, one of which is the map $$\eta:(x_1,\cdots,x_n)\mapsto (x_1,\cdots,x_n,0,\cdots,0)$$

It is natural to expect that this gives a local section of $\pi$. Lee's proof also ends with this line saying some map whose coordinate representation is $$(x_1,\cdots,x_n)\mapsto (x_1,\cdots,x_n,0,\cdots,0)$$ a local section of $\pi$.

But, I want to construct the map $V\rightarrow M$ that is a local section of $\pi$. One choice would be to consider $$\varphi^{-1}\circ \eta\circ \psi:V\rightarrow \psi(V)\subseteq \mathbb{R}^m\rightarrow \mathbb{R}^n\rightarrow U.$$ Only problem here is why would $\eta(\psi(V))\subseteq \varphi(U)$. How do I change the map so that we have composition as said above.

Assuming we have made a small change for above composition, we can see that this is actually a section (in terms of set maps for now). We need to prove that $\pi\circ(\varphi^{-1}\circ \eta\circ \psi)$ to be identity on $V$. We have $$\psi\circ(\pi\circ(\varphi^{-1}\circ \eta\circ \psi)) =(\psi\circ\pi\circ\varphi^{-1})\circ(\eta\circ \psi)$$ Let $q\in V$ and $\psi(q)=(x_1,\cdots,x_n)$. Then, $$\psi\circ(\pi\circ(\varphi^{-1}\circ \eta\circ \psi)(q)) =(\psi\circ\pi\circ\varphi^{-1})\circ(\eta\circ \psi)(q) =(\psi\circ\pi\circ\varphi^{-1})\circ(\eta(x_1,\cdots,x_n))$$ $$=(\psi\circ\pi\circ\varphi^{-1})(x_1,\cdots,x_n,0,\cdots,0) =(x_1,\cdots,x_n)=\psi(q)$$ As $\psi$ is injective, we have $$\pi\circ(\varphi^{-1}\circ \eta\circ \psi)(q)=q$$ for all $q\in V$. Thus, $\varphi^{-1}\circ \eta\circ \psi$ is a local section of $\pi$.

Any suggestions regarding how to modify $\varphi^{-1}\circ \eta\circ \psi$ is welcome. I do not understand why in the book author did not explicitly write down the map. Is it only about the existence?

  • $\begingroup$ what is the reason for downvote? $\endgroup$ – user312648 Nov 22 '17 at 18:49

Here's my solution, which is poorly alluded to in the proof in the book: Let $B_{m}(\epsilon)$ be any open ball in $\mathbb{R}^{m}$ centered at $0$ of radius $\epsilon$ such that $B_{m}(\epsilon) \subseteq \varphi(U \cap \pi^{-1}(V))$. Define $\mathcal{O}=\pi(\varphi^{-1}(B_{m}(\epsilon)))$. Notice then that since $B_{m}(\epsilon) \subseteq \varphi(U \cap \pi^{-1}(V))$, we are guaranteed by the rank theorem to have


where $B_{n}$ is the open ball in $\mathbb{R}^{n}$ centered at $0$ of radius $\epsilon$. Notice that since $\psi$ is a homeomorphism and $\psi(\mathcal{O})$ is an open set of $\mathbb{R}^{n}$, it follows that $\mathcal{O}$ is open in $V$.

Also (and here's where we address your question), given the function $\eta$ which you defined, we have that $$ \eta(\psi(\mathcal{O}))=\eta(B_{n})=B_{n} \times \{0,0,...,0\} \subseteq B_{m}(\epsilon) \subseteq \varphi(U \cap \pi^{-1}(V)). $$

Therefore, by defining $\sigma=\varphi^{-1} \circ \eta \circ \psi$ on $\mathcal{O}$, your proof goes through.


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