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In topological spaces, if $A\subset X$ and the inclusion is a homotopy equivalence, that doesn't imply that $X$ deformation retracts onto $A$. For example take one of those examples of a contractible space that doesn't deformation retract to a point.

Even in manifolds, a closed manifold doesn't deformation retract to any proper subset, one can show that with homology as done here by Olivier Bégassat.

So my question is: in the case of a manifold with boundary $(M,\partial M)$, with $N\subset \partial M$ a boundary component such that $N\hookrightarrow M$ is a homotopy equivalence, is it true that $M$ deformation retracts onto $N$?

Feel free to assume more hypothesis if needed.

Giulio
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    If everything in question is simply connected and of high enough dimension than it is even isomorphic to a cylinder by poincare duality and the h-cobordism theorem. I guess the S-cobordism theorem is a good place to look for counterexamples for non-simply connected things. – ThorbenK Mar 11 '19 at 12:42

1 Answers1

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Yes, you get a deformation retract. More formally...

Suppose $M$ is a topological manifold, $N\subseteq \partial M \subseteq M$ is a component of $\partial M$ and that the inclusion $N\rightarrow M$ is a homotopy equivalence. Then $M$ deformation retracts to $N$.

Here's the idea of the proof:

From Hatcher Algebraic Topology, Corollary 0.20 (pg. 16 in my copy), it's enough to show that $(M,N)$ has the homotopy extension property.

Now, boundary components in topological manifolds are known to have collar neighborhoods (see Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.) This is more well-known in the smooth category, where it is also much easier to prove.

So, some closed neighborhood of $N$ in $M$ is of the form $N\times [0,1] = N\times I$, where we are identifying $N$ with $N\times \{0\}\subseteq M$..

Now suppose $f:M\rightarrow X$ is a continuous function (where $X$ is any topological space). Suppose $F:N\times I\rightarrow X$ has the property that $F(n,1) = f(n)$. We wish to find a continuous function $G:M\times I\rightarrow X$ satisfying two conditions. First, for any $n\in N$, $t\in I$, that $G(n,t) = F(n,t)$. Second, for any $m\in M$, $G(m,1) = f(m)$.

To that end, set $G(m,t) = \begin{cases} f(m) & m\notin N\times [0,1]\\ F(n, \max\{s,t\}) & m=(n,s)\in N\times [0,1].\end{cases}$

To check the first condition, note that $n\in N\cong N\times \{0\}$ means the corresonding $s$ is $s= 0$. Since $t\in[0,1]$, $\max\{s,t\} = \max\{0,t\}= t$, so $G(n,t) = F(n,t).$

To check the second condition, we have $G(m,t) = f(m)$ if $m\notin N\times [0,1]$, regardless of the value of $t$. On the other hand, if $m = (n,s)\in N\times [0,1]$, then because we are taking $t=1$, we have $\max\{s,t\} = t$. So $G(m,1) = F(m,1) = f(m)$.

Finally, we must verify that $G$ is actually a continuous function. Since $f,F,$ and $\max$ are all continuous, we see that $G$ is a continuous funciton ... as long as it is a function. That is, we still need to verify that $G$ is well defined at the "transition" points where $s = 1$.

But, if $m\in N\times [0,1]$ is of the form $m = (n,1)$, then $s=1$, so $\max\{s,t\} = s=1$, so $G(m) = F(n, \max\{s,t\}) = F(n,1) = f(n).$

Jason DeVito - on hiatus
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  • +1 for pointing out that you just need to verify the homotopy extension property. – William Mar 11 '19 at 17:35
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    For the existence of collar neighborhoods of the boundary, an easier proof is also in Hatcher: proposition 3.42, p. 253. He proves this for compact manifolds but observes that the same proof works for paracompact manifolds. – Moishe Kohan Mar 12 '19 at 22:01
  • Moishe: Thanks for the reference! I had originally written this answer with $M$ a smooth manifold and realized the argument could generalize if topological manifolds with boundary had collars. I googled it, and that Annals paper came up. But it's nice to have a textbook reference! – Jason DeVito - on hiatus Mar 13 '19 at 13:40