# Composition of infinite deformation retracts

1. Jul 13, 2012

### finsly

I'm trying to give an answer to the following problem, I hope someone could come in help! Consider a smooth $n$-dimensional manifold $M$ with smooth (nonempty) boundary $\partial M$, and suppose given a function $f: M\setminus \partial M \to \mathbb{R}$ (which one can assume to be differentiable) satisfying the property that there exists $A > 0$ such that for any $A \le \alpha \le \beta$, one has that the sublevel $\left\{F\le -\beta\right\}$ is a deformation retract of $\left\{F\le -\alpha \right\}$. The question is: is it true that $\partial M$ is a deformation retract of $\left\{F\le -A\right\}\cup \partial M$ (i.e., is it true that a composition of infinitely many of such deformation retracts is a deformation retract)?

2. Jul 14, 2012

### Tinyboss

I don't have a full answer for you, but as a rule of thumb, infinite compositions of maps don't necessarily retain the properties of the individual maps. I think in this case a compactness argument might work, although I think either I'm missing something from your statement, or it's incomplete. Do we know what $f\big|_{\partial M}$ is? I was assuming it's identically zero, but I realize the problem doesn't say, nor does it say anything about what happens on the levels between zero and A.

3. Jul 15, 2012

### finsly

First of all, thank you for your reply. Next, you're right, I forgot an hypothesis that could be crucial: $f(p)\to -\infty$ as $p$ approaches the boundary $\partial M$. Could this do any difference?
Maybe, (but I don't know if this makes any sense...) an idea could be to work with the extended function $\hat{f}: M \to \mathbb{R}^*$, where $\mathbb{R}^*:=\mathbb{R}\cup \left\{\infty\right\}$ (the Alexandroff compactification of $\mathbb{R}$), $\hat{f}(p):=f(p)$ if $p \in M\setminus \partial M$ and $\hat{f}:=\infty$ if $f \in \partial M$ (hoping that this $\hat{f}$ inherits some regularity from $f$...). In this way, $\partial M$ would become the level $\left\{f=\infty\right\}$...

4. Jul 15, 2012

### lavinia

Slice the north polar ice cap off of a sphere to get a manifold with boundary. Then remove the South pole. Let f be the reciprocal of the minimum of the distances along a great circles to the South pole and to the edge of the removed polar cap. This function is continuous and f(p) -> -∞ as p approaches the edge of the removed ice cap.

But but the set,

f < - the distance of the meridian where both distances are the same

does not deform onto the edge circle of the ice cap.

it seems that you need to assume that f(p) -> -∞ if and only if p approaches the boundary.

Last edited: Jul 15, 2012
5. Jul 15, 2012

### finsly

I really apologize with all of you for the incompleteness of the provided hypothesis. Actually, the manifold $M$ is simply connected as well as its boundary $\partial M$, and these restrictions seems to exclude the latter counterexample (if I'm not wrong).
And (finally) these are all the hypothesis I have...