Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 1 '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 1 '12 at 12:49
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.