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It is written on wikipedia article (https://en.wikipedia.org/wiki/Analytic_torsion) that the Reidemeister torsion is the first invariant that could distinguish between spaces which are homotopy equivalent but not homeomorphic. Can any one give me examples of other invariants which do this?

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    $\begingroup$ That seems like a strong claim. Here is a much simpler invariant that does this: compactness. I think "spaces" here should be replaced by "closed manifolds." $\endgroup$ – Qiaochu Yuan Jun 21 '16 at 17:42
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    $\begingroup$ Or still simpler: "Being a point" distinguishes a point from $\mathbb{R}$. $\endgroup$ – David E Speyer Jun 22 '16 at 4:48
  • $\begingroup$ Dimension. Various formulations of "local dimension" apply, too. There's an enormous amount of tools. $\endgroup$ – Ryan Budney Jun 22 '16 at 5:08
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    $\begingroup$ The OP is certainly referring to distinguishing compact manifolds of the same dimension. $\endgroup$ – Mikhail Katz Jun 22 '16 at 7:45
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Usual homotopy-type invariants of the configuration spaces associated to a given space may be able to distinguish between spaces which are homotopy equivalent but not homeomorphic. For instance, Riccardo Longoni and Paolo Salvatore [Configuration spaces are not homotopy invariant, Topology 44 (2005), no. 2, 375–380; MR2114713] distinguish between the lens spaces $L(7,1)$ and $L(7,2)$ by considering the universal covers of the (ordered) configuration spaces of $2$ points on these lens spaces, and showing that a certain Massey product vanishes for $\widetilde{F_2(L(7,1))}$ but not for $\widetilde{F_2(L(7,2))}$.

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    $\begingroup$ See also this paper of Evans-Lee and Saveliev, which extends these ideas and carries out the relevant calculations to show that a few more pairs of lens spaces are not homeomorphic though they're homotopy equivalent. $\endgroup$ – Mike Miller Jun 21 '16 at 18:16
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    $\begingroup$ More generally than this there's factorization homology, which takes into account the right $E_n$-module structure on configuration spaces. $\endgroup$ – Najib Idrissi Jun 22 '16 at 10:39
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Sticking to the case of manifolds, the Kirby-Siebenmann invariant of a topological manifold $X$ is an element of $H^4(X;Z_2)$ that vanishes if $X$ admits a PL structure. (Having a PL structure is a topological invariant.) The simplest example, in some sense, is the manifold $*CP^2$ (or the Chern manifold) constructed by Freedman. It is homotopy equivalent to $CP^2$, but not homeomorphic to $CP^2$, since its Kirby-Siebenmann invariant is non-trivial.

In higher dimensions, the rational Pontrjagin classes are topological invariants by a fundamental theorem of Novikov. These can be varied more or less arbitrarily within a homotopy type, subject to the condition that the polynomial in the Pontrjagin classes (the L-polynomial) that determines the signature is constant. This is a key result of high-dimensional surgery theory; see Browder's book Surgery on Simply Connected Manifolds.

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Let

$$ X^{\square\setminus\Delta}\ :=\ \{(x\ y)\in X^2 : x\ne y\} $$

be the deleted square of X. When $\ X\ $ is a topological space, then the deleted square (with the subspace topology induced by the square) is a topological invariant which is not a homotopy invariant. Then several other derived operations are likewise topologically but not homotopically invariant even when they are a composition of the deleted square and of a homotopically invariant operation--for instance, $\ \pi_1(X^{\square\setminus\Delta})\ $ is topologically but not homotopically invariant. This worked well for manifolds (in particular in Hirsch's hands).

EXAMPLE   Let $\ I\ $ be a closed interval, and $\ T\ $ be any finite tree with more than two endpoints; thus, $\ I\ $ and $\ T\ $ are homotopically equivalent (to the 1-point space). However $\ T^{\square\setminus\Delta}\ $ is connected, while $\ I^{\square\setminus\Delta}\ $ is disconnected.

We see that the topologically invariant operation of the deleted square, as well as the fundamental group of the deleted square, can distinguish between homotopically invariant spaces (even in the simple cases).

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    $\begingroup$ I think this is essentially contained in Alex Suciu's answer (he use the notation $F_2(X)$ for the deleted square). $\endgroup$ – Mark Grant Jun 22 '16 at 8:12
  • $\begingroup$ @MarkGrant, I guess. Perhaps something like implies (not in the logical boolean sense but induces) can be even more like this. $\endgroup$ – Włodzimierz Holsztyński Jun 22 '16 at 8:22
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Intersection (co)homology is a (co)homology theory, defined for stratified pseudomanifolds that is not homotopy invariant.

Examples of applications to your question have been done by Dirk Schuetz in his papers: "Intersection homology of linkage spaces" and "Intersection homology of linkage spaces in odd dimensional Euclidean space". Using intersection cohomology he is able to distinguish non-homeomorphic spaces.

Intersection cohomology can also be used to get the topological classification of weighted projective spaces: "Intersection homology of weighted projective spaces and pseudo-lens spaces." Masato Kuwata

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