# smooth submersions and maps with local sections

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 $\pi\circ\sigma=1_U$.

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?

• what is the reason for downvote? – 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
$$\psi(\mathcal{O})=\psi(\pi(\varphi^{-1}(B_{m}(\epsilon))))=B_{n},$$
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.