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I'm related to my previous question here.

The problem is: I am given a $C^r$ manifold $M$ and a connected subset $A$ of $M$, and a $C^r$retraction $f:M\rightarrow A$ such that $f\vert_A=id:A\rightarrow A$. Then $A$ is a $C^r$ submanifold of $M$.

I have already shown that $f$ has constant rank in a neighbourhood $U$ in $M$ containing $A$. I must now use the Constant Rank Theorem, but I don't know how to. By it, there exists two diffeomorphisms, $$\alpha:U_1\underset{open}\subset U\rightarrow V\underset{open}\subset \mathbb{R}^k\times\mathbb{R}^{m-k}$$ and $$\beta:U_2\underset{open}\subset U\rightarrow W \underset{open}\subset \mathbb{R}^k\times\mathbb{R}^{m-k}$$ such that $\beta\circ f \circ \alpha^{-1}(x,y)=(x,0)$.

Now, what should be the next step to prove that $A$ is a submanifold? I tried working with the projection $$ \pi_2:\mathbb{R}^k\times\mathbb{R}^{m-k}, \pi_2(x,y)=y $$ but couldn't use the regular value theorem. Can someone help me?

Community
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Marra
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  • Doesn't $\beta$ give you a "slice chart" for $A$? By slice chart I mean a chart for the ambient manifold that takes $A$ to a linear subspace. The existence of a slice chart at every point of A implies A is an imbedded submanifold. The immediacy of the implication depends on your definition of submanifold. But maybe I am misunderstanding your question. – Tim kinsella Aug 24 '13 at 05:04
  • Yes,but I guess that $\beta$ is a slice chart only when associated with the chart "\alpha" on the other side, right? In this situation, my question would be "why slice chart at every point implies submanifold"? – Marra Aug 24 '13 at 13:43
  • What is your definition of submanifold? – Tim kinsella Aug 24 '13 at 17:14
  • I am using Morris' "Differential Topology" as reference. – Marra Aug 26 '13 at 16:13

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I think your argument is complete:)

First of all, one of the definitions of submanifold is "locally image of immersion".

Secondly, regular value theorem is actually a special case of the constant rank theorem ($\beta$ is not needed in r.v.t. but actually it is a very simple function [no IFT magic, an honest formula]) and you don't need to use it (r.v.t. $\simeq$ c.r.t.) twice.

Just have a look: locally - in chart $\beta$ - the subset $A$ is a linear subspace of dimension $k$ (note that your formula [c.r.t] guarantees that $A\cap U_2$ is exactly this subspace) and since a linear subspace is obviously image of immersion (even a linear monomorphism), it is a submanifold.

savick01
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  • Isn't it the image of a embedding? – Marra Sep 04 '13 at 20:29
  • @GustavoMarra What do you mean by embedding? Just a smooth homeomorphism onto the image? It doesn't have to be a submanifold. – savick01 Sep 04 '13 at 20:43
  • I meant that one of the definitions ofa submanifold is "image of a embedding". – Marra Sep 04 '13 at 20:45
  • That is a very strange definition. I'd even say incorrect. Consider embedding $f:\mathbb R \to \mathbb R^2$ given by $f(x)=(\sqrt[3]{x},0)$. As a set, the image of course is a submanifold - however smooth functions on $\mathbb R^2$ do not restrict to smooth functions on $\mathbb R$ parametrized by $f$. – savick01 Sep 04 '13 at 20:52
  • It's on Hirsch's "Differential Topology" at page 21, Theorem 3.1 – Marra Sep 04 '13 at 20:54
  • @GustavoMarra Is it somewhere online? – savick01 Sep 04 '13 at 21:00
  • Not that I know of; but the theorem states that A subset $A\subset M$ is a submanifold (M manifold of class $C^r$) if and only if $A$ is the image of a $C^r$ embedding. – Marra Sep 04 '13 at 21:03
  • @GustavoMarra But what is embedding for you? On wikipedia the definition agrees with mine, but is formulated with the word "embedding" (which is required to be an immersion). – savick01 Sep 04 '13 at 21:06
  • An embedding is an immersion which is a homeomorphism over its own image. – Marra Sep 04 '13 at 21:10
  • @GustavoMarra OK, so our definitions agree:) Immersion is locally a homeomorphism onto its image and that's what I meant. Formally your definition is global, but it doesn't matter. – savick01 Sep 04 '13 at 21:11
  • They agree due to the Inverse/Implicit Function Theorem, right? – Marra Sep 04 '13 at 21:15
  • @GustavoMarra Yes. – savick01 Sep 04 '13 at 21:17
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    Thank you! Here is your reward! :-P – Marra Sep 04 '13 at 21:18