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What is the difference between homotopy and homeomorphism? Let X and Y be two spaces, Supposed X and Y are homotopy equivalent and have the same dimension, can it be proved that they are homeomorphic? Otherwise, is there any counterexample? Moreover, what conditions should be added to homotopy to get homeomorphism?

We assume additionally both X and Y are orientable.

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Nitpick: Maps are homotopic. Spaces are homotopy equivalent. –  Neal Jan 18 '13 at 12:27
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When you say $X$ and $Y$ are homotopic, I assume you mean that they are homotopy equivalent. Anyways, homotopy equivalence is weaker than homeomorphic.

Counterexample to your claim: the 2-dimensional cylinder and a Möbius strip are both 2-dimensional manifolds and homotopy equivalent, but not homeomorphic.

Unfortunately I'm not an expert on the subject so I'm not sure what are the weakest assumptions to add to homotopy to get a homeomorphism.

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The counterexample is right. While what about if we assume additionally both X and Y are orientable? –  liufu Jan 18 '13 at 12:28
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@liufu: technically orientiability has no meaning for arbitrary spaces. If you restrict to spaces like manifolds where such ideas are well-defined then the answer is still no. Sigur's lens space example is one. –  Ryan Budney Sep 11 '13 at 22:46
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Simplest counter-example is $X = B_n(0)$ (a ball of dimension $n$ around $0$) and $Y = \{0\}$ with the contraction $H(t,x) = tx$, i.e. $H(0,x) = x$, $H(1,x) = 0$. You will not find a homeomorphism though, since the sets don't even have the same cardinality.

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This example is not valid here, since the OP asked for examples with the same dimension. –  Per Manne Jan 18 '13 at 12:22
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Look for 3-dimensional lens spaces of type $L(p,q)$, quotient of $S^3$ by a free orthogonal action of the cyclic group $\mathbb{Z}_p$. More precisely, look for the spaces $L(5,1), L(5,2)$ and $L(7,1), L(7,2)$.

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I you don't necessarily want compact manifolds, in dimension $3$ and up there are manifolds which are contractible (i.e., homotopy equivalent to a point), but not homeomorphic to$\mathbb{R}^n$. See, for example, the Whitehead manifold.

I you insist on compact simply connected manifolds, I don't know of too many examples. For example, in Kamerich's thesis "Transitive transformation groups on products of two spheres", he proves

The homogeneous space $(Sp(24)\times Sp(2))/(Sp(23)\times \Delta Sp(1) \times Sp(1))$ given by the embedding $(A,p,q)\mapsto \big(\operatorname{diag}(A,p), \operatorname{diag}(p,q)\big)$ where $Sp(n)$ denotes $n\times n$ quaternionic unitary matrices is homotopy equivalent but not homeomorphic to $S^{95}\times S^4$. Further, this is best result possible in the sense that replacing $24$ and $23$ with smaller numbers always gives examples which are not homotopy equivalent to any product of spheres.

Onishik's book "The Topology of Transformation Groups" pg. 275 contains more details.

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Nice book, thanks. –  Sigur Jan 18 '13 at 14:03
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This is the content of certain rigidity theorems.

You should check out the Mostow rigidity theorem. It implies, that given two smooth closed manifolds which are homotopy equivalent and both hyperbolic (constant sectional curvature = -1) then they are diffeomorphic (see http://en.wikipedia.org/wiki/Mostow_rigidity_theorem ).

Note that by the Cartan-Hadamard theorem it follows that a hyperbolic manifold is what is called aspherical, i.e. its universal cover is contractible.

There is a very beautiful conjecture due to Borel (the Borel Conjecture) that can be phrased as follows.

Let $f: M \to N$ be a homotopy equivalence of closed aspherical manifolds. Then $f$ is homotopic to a homeomorphism, and in particular $M$ and $N$ are homeomorphic.

Note that this conjecture assumes less about the manifolds (every hyperbolic manifold is aspherical, but not every aspherical manifold is hyperbolic) but you also get a weaker conclusion (the manifolds are homeomorphic not diffeomorphic or isometric).

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For hyperbolic manifolds (of finite volume), Mostow rigidity is seemingly even stronger. Namely, any two such manifolds with isomorphic fundamental group are isometric. (Since hyperbolic manifolds are $K(\pi, n)s$, this isn't really stronger, but it certainly seems stronger on first notice). –  Jason DeVito Jan 18 '13 at 16:56
    
Yes, but I did not want to mention it in this way in order to stay close to the question :) –  mland Jan 18 '13 at 19:45
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Let $X$ be the letter

$$\ \ \ \ \ \mathsf{X}\ \ \ \ \ $$ and $Y$ be the letter

$$\ \ \ \ \ \mathsf{Y}\ \ \ \ \ $$

Then $X$ and $Y$ are homotopy-equivalent, but they are not homeomorphic.


Sketch proof: let $f:X\to Y$ map three of the prongs of the $\mathsf{X}$ on to the $\mathsf{Y}$ in the obvious way, and let it map the fourth prong to the point at the centre. Let $g:Y\to X$ map the $\mathsf{Y}$ into those three prongs of the $\mathsf{X}$. Then $f$ and $g$ are both continuous, and $f$ is a surjection but is not injective, while $g$ is an injection but is not surjective. Now the compositions $f\circ g$ and $g\circ f$ are both easily seen to be homotopic to the identities on $X$ and $Y$, so $X$ and $Y$ are homotopy-equivalent.

On the other hand, $X$ and $Y$ are not homeomorphic. For example, removing the point at the centre of the $\mathsf{X}$ yields a space with four connected components, while removing any point from the $\mathsf{Y}$ yields at most three connected components.

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