# (Whitney) Extension Lemma for smooth maps

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.

## 1 Answer

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!

• 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? – Alex Ortiz Mar 13 '18 at 17:56
• @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 $[0,\infty)$. – Jack Lee Mar 13 '18 at 18:30