10
$\begingroup$

I am currently reading Lee's book "Introduction to Smooth Manifolds (2nd edition)".

Corollary 6.27 in that book states that a smooth map $f\colon A \rightarrow M$ where $M$ is a smooth manifold without(!) boundary and $A \subset N$ is closed (where $N$ is a manifold with possibly nonempty boundary) can be extended to a smooth map $F\colon N \rightarrow M$ iff it has a continuous extension.

Lee claims (and leaves this as Problem 6-7), that the claim is in general false if $M$ has nonempty boundary. The Problem asks you to show this by considering $F\colon \mathbb{R} \rightarrow \mathbb{H}^2, t \mapsto (t, |t|)$ with $\mathbb{H}^2 = \mathbb{R} \times [0,\infty)$ and $A = [0,\infty)$.

The problem with this Problem is that Lee defines a map $f\colon A \rightarrow M$ on a subset $A \subset N$ to be smooth if for every $p \in A$ there is an (open) neighborhood $W_p \subset N$ of $p$ and a smooth map $f_p\colon W_p \rightarrow M$ that agrees with $f$ on $A \cap W_p$.

This condition is clearly not satisfied for the above $F$ at $t = 0$.

My question is, whether the claim nevertheless fails (with another counterexample).

The problem in instead proving the claim in this case is that Lee uses an embedding of $M$ in $\mathbb{R}^n$ (no problem) and then uses the existence of a "tubular neighborhood" which fails if $M$ has nonempty boundary.

Any help would be appreciated.

$\endgroup$
19
$\begingroup$

Uh-oh ... this is a flaw in my definition of smoothness of a map defined from a subset of a manifold (with or without boundary) into a manifold with boundary. If $M=\mathbb R$, $N=[0,\infty)$, and $A = [0,\infty)\subset M$, we would certainly want to consider the map $f\colon A\to N$ defined by $f(x)=x$ to be smooth, but it's not by the definition I gave. I've added a revised definition to my list of corrections (available here; see the correction to p. 45) that fixes this problem.

Good catch!

$\endgroup$
  • $\begingroup$ Let $W = (-1,1)$, let $V = [0,1)$, $\psi = \mathrm{Id}_{\Bbb{R}_{\ge 0}}$. Then we have that $F(W\cap A) = F((-1,1)\cap[0,\infty)) = F([0,1)) = [0,1) \subset V$, but there does not appear to be any smooth extension of $g \equiv \psi\circ F|_{W\cap A}$. For, since the codomain of $g$ is $[0,\infty)$, if $\widetilde g$ were an extension, then necessarily $\widetilde g(0) = 0$ is a (global) minimum for $\widetilde g$, so $\widetilde g'(0) = 0$. However, by continuity, $\widetilde g'(0) = \lim_{x\to0^+}\widetilde g'(x) = \lim_{x\to 0^+} g'(x) = 1$, which is a contradiction. Is the definition okay? $\endgroup$ – Alex Ortiz Mar 13 '18 at 17:56
  • 1
    $\begingroup$ @AOrtiz: The correction says that $\psi\circ F|_{W\cap A}$ has to be smooth as a map into $\pmb{\mathbb R^n}$. The whole point of the correction is that the extension doesn't have to take its values in $[0,\infty)$. $\endgroup$ – Jack Lee Mar 13 '18 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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