9

Looking over some old qualifying exams, we found this:

Let $A\subseteq M$ be a connected subset of a manifold $M$. If there exists a smooth retraction $r:M\longrightarrow A$, then $A$ is a submanifold.

Our thought to prove this statement was that since $r$ is smooth and the identity on $A$, then the inclusion $i:A\longrightarrow M$ is smooth. Also, since $i\circ r=\operatorname{Id}_A$, then $i_*:TA\longrightarrow TM$ is injective. Thus $i$ is a smooth immersion. Therefore $A$ is a submanifold. But, nowhere did we use that $A$ is connected. What is wrong with the argument? And, what is the correct proof?

J126
  • 17,451
  • How to establish the coordinate charts of $A$? And what is the definition of smooth mapping $i:A\rightarrow M$? Since here you haven't given $A$ a differentiable structure. – Yuchen Liu Aug 01 '12 at 12:18
  • Dear Joe, your reasoning is circular : what does $TA$ mean if you don't know that $A$ is a manifold? Similarly, what does "immersion" mean ? – Georges Elencwajg Aug 01 '12 at 12:49
  • I think the OP saw this exercise in Hirsch's book Differential Topoplogy (Chapter 1, sec. 2 exercise 2). There, $A$ is assumed to be a connected manifold in its own and the retraction $f:M\rightarrow A$ is a $C^r$-map. Then, one has to show that the smooth structure on $A$ is precisely the one inherited from $M$ (at least that's how I see it). – Dog_69 Oct 18 '19 at 14:45
  • Here I post the Exercise: Let $M$ be a $C^r$ manifold, $r\geq 1$, and $A\subset M$ a connected subset. Suppose that there is a $C^r$ retraction $f:M\rightarrow A$, i.e. $f|A=id$. Then $A$ is a $C^r$ submanifold. (The converse is proved in Chapter 4.) [Hint: $f$ has constant rank near $A$.] – Dog_69 Oct 18 '19 at 14:49

1 Answers1

6

As pointed out in the comments, since you don't know $A$ is a manifold, you can't speak about smooth immersion of $A$ into $M$.

To prove the statement, you have to show there exists an open neighborhood $U$ of $A$ in $M$ such that the rank of $T_y r$ is constant for $y\in U$. Then applying the constant rank theorem, the result follows.

If $A$ was not connected, in general, the rank of $T_y r$ would have a different value in each connected component and $A$ would not be a pure manifold.

For the proof details you can look at P. W. Michor, Topics in Differential Geometry, section 1.15.

AlbertH
  • 5,338
  • 1
    How exactly does one use the constant rank theorem? I understand the construction of the neighborhood U. I just do not get how to use the constant rank theorem to construct the submanifold charts for A.

    For every $p \in U$, the constant rank theorem gives me charts $(V,v)$ and $(W,w)$ centered around $p$ and $r(p)$, respectively, such that $\hat{r} = w \circ r \circ v^{-1}$ looks locally like $\hat{r}(x,y) = (x,0)$ for all $(x,y) \in v(U)$.

    Is the idea to use $(V,v)$ as submanifold charts around $r(a) = a$ for all $a \in A$?

     – Jan Vysoky May 07 '19 at 21:50