# Is it possible for a closed manifold to deformation retract onto a proper subset of itself?

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 $A$?

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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.

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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