Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M$ be a closed (compact, without boundary) topological manifold. Is it possible for there to exist a subset $A$ of $M$ such that $M$ deformation retracts onto $A$?

share|improve this question

1 Answer 1

up vote 11 down vote accepted

I'm not 100% sure of my answer. I'll assume $M$ to be connected; $n$ stands for its dimension. Suppose $M$ deformation retracts onto a proper subet $A$.

Because $M$ deformation retracts onto $A$, the long exact homology sequence (LEHS) of the triple $\emptyset\subset A\subset M$ with coefficients in $G=\Bbb Z$ or $G=\Bbb Z/2\Bbb Z$ (depending on wether $M$ is orientable or not) tells us $$H_*(M,A)\equiv 0.$$

Using the above and the LEHS for $A\subset M\setminus\text{pt}\subset M$, there are isomorphisms $$H_{*-1}(M\setminus\text{pt},A)\simeq H_*(M,M\setminus\text{pt})\simeq H_*(\Bbb R^n,\Bbb R^n\setminus 0).$$ The first one is the connecting homomorphism, the second one arises from excision. Therefore $H_*(M\setminus\text{pt},A)=0$ for $*=n,n+1$ Then, the LEHS of the triplet $\emptyset\subset A\subset M\setminus \text{pt}$ in degree $n$ tell us that

$$H_n(A)\simeq H_n(M\setminus \text{pt})$$

Since $M$ is closed connected, its top homology is isomorphic to $G$, and so is that of $A$ since they are homotopy equivalent: $$H_n(M)\simeq H_n(A)\simeq G.$$ Also, since $M\setminus\text{pt}$ is a non compact connected manifold, its top homology is $0$: $$H_n(M\setminus\text{pt})=0.$$

This contradicts the isomorphism $H_n(A)\simeq H_n(M\setminus \text{pt})$. Thus there are no deformation retractions of $M$ onto a proper subset.

share|improve this answer
A formally essentially identical argument but slightly more direct is to observe that the degree of the identity map is $1$. But if the manifold is homotopy-equivalent to a proper subset, the identiy map would have degree zero. –  Ryan Budney Dec 14 '12 at 3:41
Thanks! Just one question: how do we know that the top homology of a non-compact connected manifold is 0? –  user15464 Dec 14 '12 at 3:51
@user15464 I don't remember the argument, but I believe it's a theorem in Milnor and Stasheff's Characteristic Classes. If I'm not way out of line, I think it's got to do with the fact that any cycle will have compact support, but I don't recall the argument. –  Olivier Bégassat Dec 14 '12 at 3:56
@RyanBudney Is this because a map that misses a point will have degree zero? I know the argument for maps from $\Bbb S^n$ to itself (stereographic projection), but why does it hold (if indeed it does) for compact manifolds other than the sphere? –  Olivier Bégassat Dec 14 '12 at 4:01
There's a formula for computing the degree as a signed count of points in the pre-image of a regular value. A point not in the image is a regular value. –  Ryan Budney Dec 14 '12 at 4:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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