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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?

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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
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.

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