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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 know that I need to show that $f$ has constant rank in a neighbourhood $U$ (open in $M$) of $A$ so I can apply the Rank Theorem to finish the proof. I tried the same approach as in P. W. Michor, Topics in Differential Geometry, section 1.15, but since the hypothesis in this book is kinda different I couldn't use it to get the result I need.

I thought that since $f\vert_A=id:A\rightarrow A$, then for every $x\in A$, $T_xf=Id$ which is surjective. Therefore $f$ is a local submersion in $x$ and so, in a neighbourhood $U_x\underset{open}\subset M$ $f$ has constant rank. Now I can collect all $U_x$ and get an open set $U$ in $M$ where $f$ has constant rank.

Is this correct?

By the way, this is question 2, Section 2 Chapter 1 from Hirsch's "Differential Topology".

Thanks.

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Marra
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  • I think you might need to be more careful about the boundary points $x \in A \cap \partial A$. – Anthony Carapetis Aug 23 '13 at 04:21
  • Could you be more specific? If $x$ is in this set the $x\in A$. – Marra Aug 23 '13 at 04:24
  • Is it immediate that $T_x f =\operatorname{Id}$ in this case? Thinking about it a little more I don't think there's an issue so long as $f$ is $C^r$. – Anthony Carapetis Aug 23 '13 at 04:28
  • I forgot to mention that $f$ is also $C^r$ as a mapping. – Marra Aug 23 '13 at 04:30
  • Maybe I should break it in two cases; the first one being when $A$ is open in $M$ (this is trivial) and the other where $A$ is not open. This is where $\partial A$ comes into play, if $A$ is closed then I guess it's OK (for $Df(x)=Id$ is guaranteed in $\partial A$) but if it's not there might be problems. – Marra Aug 23 '13 at 04:37
  • But if $A$ is not closed, if there's a point $X$ in $\partial A$ where $f(x)\neq x$, then there is a neighbourhood of $x$ where $f(y)\neq y$ in this neighbourhood, therefore there would be a point $w$ in $A$ such that $f(w)\neq w$, contradiction. Is this OK? – Marra Aug 23 '13 at 04:41
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    Notice that $A$ is exactly the fixed point set of $f:M \to M$. Your argument then works; alternatively just observe that fixed point sets of continuous maps of Hausdorff spaces are closed. – Anthony Carapetis Aug 23 '13 at 05:05
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    What's wrong with just quoting Michor's result? The fact that $f$ is a retraction implies $f\circ f=f$. Here's the problem with your approach: If you are considering $f$ as a map from $M$ to $A$, then $T_xf$ does not make sense, because $A$ is not yet known to be a manifold. On the other hand, if you consider $f$ as a map from $M$ to $M$ (whose image happens to be $A$), then it's not generally true that $T_xf = \operatorname{Id}$, because $f$ will not generally be equal to the identity on an open set. – Jack Lee Aug 24 '13 at 17:52
  • @JackLee I see the problem in my approach now. Thanks! I'll try to use the same argument as Michor, since $f\circ f=f$ in $M$. – Marra Aug 26 '13 at 15:18

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