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In algebraic topology, we have the definition of homotopy equivalence of topological spaces:

Topological spaces $X$ and $Y$ are homotopy equivalent if there exist continuous maps $\phi:X \rightarrow Y$ and $\psi:Y\rightarrow X$ such that $\psi \circ \phi \simeq Id_X$ and $\phi \circ \psi \simeq Id_Y$.

If you compare this with the definition of homeomorphism, there are various differences.

However, in terms of finding examples where two spaces are homotopy equivalent and not homeomorphic, I struggle. The only examples I can find are the cases where we involve 1-point sets. For example, a disk in $\mathbb{R}^2$ centred at the origin is not generally homeomorphic to a 1-point set, but it is homotopy equivalent to, say, the origin.

The motivation for this question is that if I were asked whether two spaces were homotopy equivalent - and neither are a 1-point set, could I just check whether they were homeomorphic?

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    $\begingroup$ An instructive exercise is to classify the letters of the alphabet up to homeomorphy and up to homotopy equivalence. $\endgroup$ – tschih Jan 2 '18 at 1:44
  • $\begingroup$ $X$ is always homotopy equivalent to $X \times \mathbb{R}$ but usually not homeomorphic (say if $X$ is a manifold or finite CW complex, for dimension reasons). $\endgroup$ – Qiaochu Yuan Jan 2 '18 at 1:53
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One non-trivial example is that the Möbius strip is homotopy equivalent to a cylinder. These spaces are not homeomorphic, since the boundary of a cylinder (as a manifold) consists of two circles, whereas the boundary of a Möbius strip consists of just one.

To see that they are homotopy equivalent, it suffices to note that they are both homotopic to a circle. One can see this by noting that we can deformation retract either figure "linearly" to a circle down the middle, by pulling the boundaries together.

We can also see that some general constructions hold to find pairs of homotopy equivalent spaces. In particular, if $X$ and $X'$ are homotopy equivalent, as are $Y$ and $Y'$, then so are $X\times X'$ and $Y\times Y'$. In particular, this implies that if a space $Y$ is contractible, then $X\times Y$ is homotopic to $Y$.

One can also find that if the pairs $(X,x)$ and $(X',x')$ are homotopy equivalent and $(Y,y)$ and $(Y,y')$ are homotopy equivalent, then $X\vee Y$ is homotopy equivalent to $X'\vee Y'$, where $\vee$ is the wedge sum. In a sense, once we have a few examples of things being homotopy equivalent, any space we can build out of them becomes homotopy equivalent too - so contractible spaces are a whole class that homotopy equivalence cannot see.

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  • $\begingroup$ Ah, another great example. $\endgroup$ – Randall Jan 2 '18 at 1:40
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    $\begingroup$ Nice examples. But I'd suggest that you replace "homotopic" with "homotopy equivalent" throughout, reserving "homotopic" for pairs of maps. I know some folks does use "homotopic" this way, but I think that keeping them distinct is useful, esp. for beginners. $\endgroup$ – John Hughes Jan 2 '18 at 1:59
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Homotopy equivalence is absolutely a weaker notion than being homeomorphic. The latter is incredibly strong.

Take the unit circle $S^1$, and then compare to the unit circle with a whisker* attached. By retracting the whisker they are homotopy equivalent. But they are not homeomorphic: cutting any point in $S^1$ leaves a connected space, but this is not true for the circle with a whisker.

*If you've not seen this before, you can build this space as the union of $S^1$ with the interval $[1,2]$ on the $x$-axis.

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In this context a mathematical joke would be: Give an example of two spaces $X,Y$ which are homotopically equivalent but not homeomorphic.

Observe that if two spaces are homeomorphic, then they are of same dimension, whereas for homotopy equivalent they need not be of same dimension.(in same way you don't need compactness assumption). Intuitively two spaces are homotopically equivalent means there exists a bigger space which is deformation retract onto both spaces.

One better question is if they are of same dimension, homotpically equivalent, and compact, does that mean they are homeomorphic. Still the answer is no. Any two lense spaces $L(p,q)$ with different $q (mod p)$ are homotopically equivalent but non-homeomorphic.

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    $\begingroup$ You seem to consider only manifolds; there is no standard notion of dimension for arbitrary topological spaces. $\endgroup$ – Danu Jan 2 '18 at 13:02
  • $\begingroup$ @danu I'm mostly interested in cw- complexes...and there dimension is well defined i.e, the dimension of top dimensional face ( connected) $\endgroup$ – Anubhav Mukherjee Jan 2 '18 at 14:42
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A hard version of this question is to restrict attention to closed manifolds: for example, it's in fact the case that two closed $1$-manifolds or two closed $2$-manifolds are homotopy equivalent iff they are homeomorphic. In higher dimensions we also have Mostow rigidity, which guarantees that closed hyperbolic manifolds are homotopy equivalent iff they are homeomorphic.

The smallest counterexamples are closed $3$-manifolds, namely the lens spaces: these are a family of closed $3$-manifolds $L(p, q)$ defined by a pair of coprime integers, with fundamental group $C_p$. It's known that $L(p, q_1)$ and $L(p, q_2)$ are

  • homotopy equivalent iff $q_1 q_2 \equiv \pm n^2 \bmod p$ for some integer $n$, and
  • homeomorphic iff $q_1 \equiv \pm q_2^{\pm 1} \bmod p$.

It follows that the lens spaces $L(7, 1)$ and $L(7, 2)$ are homotopy equivalent (because $1 \cdot 2 \equiv 3^2 \bmod 7$) but not homeomorphic (because $1 \not \equiv \pm 2^{\pm 1} \bmod 7$). This is quite a hard example because $L(7, 1)$ and $L(7, 2)$ can't be distinguished by any of the usual invariants from general or algebraic topology; distinguishing them requires a new invariant called Reidemeister torsion.

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Not really, no. Any deformation retraction defines a homotopy equivalence between two spaces, and this is a pretty good way to generate lots of examples that don’t reduce to the 1-point case.

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