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 $A$?
Is it possible for a closed manifold to deformation retract onto a proper subset of itself?
1 Answers
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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0@Bruno given a triple $(X,A,B)$ (that is $X\supset A\supset B$) there are connecting homomorphisms $\partial=\partial_{X,A,B,n}:H_{n+1}(X,A)\to H_n(A,B)$ for a long exact homology sequence: $\cdots\stackrel{\partial}{\to}H_{n+1}(A,B)\to H_{n+1}(X,B)\to H_{n+1}(X,A) \stackrel{\partial}{\to}H_{n}(A,B)\to\cdots$ where all the other maps are induced by the inclusions of pairs. This follows from the long exact homology sequence defined by a short exact sequence of complexes, here $0\to C_*(A,B)\to C_*(X,B)\to C_*(X,A)\to 0$, where the $C_*(A,B)$ are the relative singular chain complexes. – 2013-09-27