# Is there a general way to tell whether two topological spaces are homeomorphic?

We know that if two topological spaces $X$ and $Y$ are homeomorphic, then they have the same fundamental groups, and the same homology. In other words, we have functors $$\pi_1 : \mathsf{Top} \to \mathsf{Grp} \quad\text{and}\quad H_n : \mathsf{Top} \to \mathsf{Ab}$$ (actually this works even if the spaces are homotopy equivalent). The important thing here is that these functors can be used to prove that the two spaces are not homeomorphic: for instance $H_3(S^3) \cong \Bbb Z \not\cong 0 = H_3(S^2)$, so that $S^3$ and $S^2$ are not homeomorphic (they don't even have the same homotopy type).

I was wondering whether there was somehow a "converse" to this, i.e. is they a way to prove that two topological spaces are homeomorphic. More precisely:

Is there a category $\scr C$ and a functor $\mathsf F : \mathsf{Top} \to \mathscr C$ such that $\mathsf F(X) \cong \mathsf F(Y) \implies X \cong Y$ ? Of course, I want to avoid obvious examples as $\mathsf{Id_{Top}}$ .

(By the way, I don't know if there is a name for such functors, which are injective on objects. Faithful is already used for something different). I would also accept discussing the case where the homeomorphism $X \cong Y$ is replaced by a homotopy equivalence $X \simeq Y$.

Trying the functor $\mathsf F(X) = X \times X$ doesn't work, as shown here. The functor $\mathsf F(X) = X \sqcup X$doesn't work as well.

The closest result I found is a theorem due to Gelfand and Kolmogorov : given two compact and Hausdorff spaces, if the commutative rings $C(X)$ and $C(Y)$ of continuous functions $f\,:\,X,Y\rightarrow \mathbb{R}$ (under pointwise addition and multiplication) are isomorphic, then $X$ and $Y$ are homeomorphic. Maybe we could try to generalize this to the category of locally compact and Hausdorff spaces, using the Alexandorff compactification.

• Invariants are useful for telling spaces apart, as you have shown, but not to show they are isomorphic. For example, even cohomology algebras can fail to detect attachings of a space onto another by different homotopy clases of maps. – Pedro Tamaroff Jul 24 '16 at 9:25
• what properties on $\mathscr C$ do you want to hold? for example, the category of motives (universal cohomology theory for algebraic manifolds) desired to be abelian. please, specify in what sense $\mathscr C$ must be better then $\mathrm{Top}$? you just want to change the notion of equivalence in first category to another one, so $\mathscr C$ must be more convenient to work with. – Andrey Ryabichev Jul 24 '16 at 13:04
• Regarding your terminology question: functors that reflect isomorphisms (ie. $Ff$ iso implies $f$ iso) are called conservative. They can detect if a given morphism is iso, but they can't tell you if an isomorphism exists. An example is $U$ : Grp → Set, counterexample $U$ : Top → Set. The property you want is $FX ≅ FY$ implies $X ≅ Y$, and I don't know if it has a name, although "weakly injective on objects" seems like a reasonable candidate. – user54748 Jul 24 '16 at 13:16
• Perhaps Whitehead's theorem deserves mention in this context: If we have a map $f:X\to Y$ between simply connected CW complexes, which is an isomorphism on $\pi_*$, then we can conclude that $X\simeq Y$. – iwriteonbananas Jul 24 '16 at 13:42
• For your reference, there exists the theorem of Freedman in 4-dimensional topology : the homeomorphic type of simply-connected closed smooth 4-manifold is determined by the intersection form. – Shinichiro Nakamura Jul 25 '16 at 3:04

No, at least in the sense that there are no such functors $F$ which are substantially more approachable than the original problem. There are very few general techniques for distinguishing homotopy equivalent, non-homeomorphic spaces: see the following link for work in the last twelve years distinguishing some such pairs of 3-dimensional closed manifolds, which are a ludicrously tiny tip of the iceberg of arbitrary spaces. It's worth noting here, to give a sense of how far from possible a simple, complete homeomorphism invariant is, that even the homotopy equivalence problem runs into uncomputability, insofar as it's undecidable whether two spaces have the same fundamental group, given presentations of those groups. https://mathoverflow.net/questions/242748/list-of-invariants-that-distinguish-homotopy-equivalent-non-homeomorphic-spaces

• Thank you very much! "See the following link"… I don't see any link ;-) – Watson Jul 25 '16 at 20:13
• Oops! It's there now. – Kevin Carlson Jul 25 '16 at 20:38