Is defining map always a submersion? Why?

Question

Is local defining map always a submersion? Why?

I came up with this issue in some part of my proof for other things.. This might be a stupid question, but thank you guys for your help in advance.

Answer 1

You can probably answer your own question if you just work backwards through the definitions in the book. On page 107, I define a local defining map for a submanifold $S\subseteq M$ as "a smooth map $\Phi\colon U\to N$ such that $S\cap U$ is a regular level set of $\Phi$" (where $U$ is an open subset of $M$). Then if you look back at the top of page 106, I defined a regular level set of $\Phi$ to be a subset of the form $\Phi^{-1}(c)$ (shorthand for $\Phi^{-1}(\{c\})$) where $c$ is a regular value of $\Phi$. A little before that, I said "$c$ is a regular value of $\Phi$ if every point of the level set $\Phi^{-1}(c)$ is a regular point." And back a few lines further, "a point $p\in M$ is said to be a regular point of $\Phi$ if $d\Phi_p\colon T_pM \to T_{\Phi(p)}N$ is surjective."

Once you've found all those definitions, I recommend that you go back to the book and read them in forward order, taking some time to absorb each definition as you read it. There are a lot of definitions being layered on top of each other here, so you shouldn't feel bad if you don't instantly grasp them the first time you read them.

The upshot is that what's being assumed when we talk about a "local defining map" is that $d\Phi_p$ is surjective at each point $p\in S\cap U$. To say $\Phi$ is a submersion would mean that $d\Phi_p$ is surjective at every point $p$ in the domain of $\Phi$. So the answer is no, a local defining map need not be a submersion.

Here's a simple counterexample. Let $S$ be the unit circle in $\mathbb R^2$. The map $\Phi\colon \mathbb R^2\to \mathbb R$ given by $\Phi(x,y) = x^2 + y^2$ is a local defining map for $S$ (actually a global one in this case, but every global defining map is also a local defining map). But it's not a submersion because its differential vanishes at the origin.