What is the difference between homotopy and homeomorphism?

Solution 1:

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.

In other words, observe that $\mathsf Y$ is a deformation retract of $\mathsf X$. Alternatively, observe that $\mathsf X$ and $\mathsf Y$ both retract on to the point at the centre.

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.

Solution 2:

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.

Solution 3:

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).

Solution 4:

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 $L(7,1), L(7,2)$.

Solution 5:

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.

Onishchik's book "Topology of Transitive Transformation Groups" pg. 275 contains more details.