A while ago a professor of mine said something along the lines of

No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants

Where algebraic invariants are functors from $Top$ to $Grp$ I guess. I've not been able to find this result anywhere online (partly because I don't know what to google), so I'm looking for a source where this result is discussed.

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    $\begingroup$ This question is in my opinion really vague. What is the precise definition of "algebraic invariant"? If for instance you mean a functor from $\mathsf{Top}$ to $\mathsf{Grp}$ which is homotopy invariant (a very reasonable request), then any two non-homeomorphic, homotopy equivalent spaces cannot be distinguished by any invariant. $\endgroup$ Mar 23, 2016 at 9:05
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    $\begingroup$ @FrancescoPolizzi Of course, if you restrict yourself to only homotopy invariant functors the question is trivial, hence I presume the result would be about all functors (homotopy invariant or not). I'm sorry if the question is vague, this statement was only mentioned in passing so I don't know the exact formulation; I was just hoping someone would recognise it and point me to a source. $\endgroup$ Mar 23, 2016 at 9:11
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    $\begingroup$ This idea is not so precise, but here you go. Choose some skeleton of Top, and some skeleton of Grp, and construct a bijection between them (I guess these have the same "cardinality"). Map identities to identities, and all functions to zero homomorphisms. Extend this to all topological spaces. Now you have your -not so useful- algebraic invariants that distinguish spaces up to homeomorphism.. $\endgroup$
    – Thomas Rot
    Mar 23, 2016 at 9:30
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    $\begingroup$ The proper (so-to-speak) version of the problem is: *** there doesn't exist any algorithm which decides about two spaces whether or not they are homeomorphic. *** Here we can actually restrict our attention to finite polyhedra only. $\endgroup$ Mar 23, 2016 at 9:46
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    $\begingroup$ @WłodzimierzHolsztyński the algorithmic problem is distinct. A faithful functor frop Top to Grp would not yield an algorithm in principle. $\endgroup$
    – YCor
    Mar 23, 2016 at 10:09

3 Answers 3


perhaps he is implying some even stronger result

He is referring to the following result of Peter Freyd (Freyd uncertainty principle):

The homotopy category of spaces $HoTop$ does not admit a faithful functor to the category of sets $Set$. Specifically, for any functor $T: Top_* \to Set$ from base-pointed spaces to sets which is homotopy invariant, there exist a triple $f: X \to Y$ such that $f$ is not null-homotopic, but $T(f) = T(\ast)$. Here $\ast$ is the null map to the basepoint of $Y$.

In particular, any algebraic invariant is a set-valued homotopy invariant. This includes homotopy groups, cohomology, cohomology and homotopy operations and whatever you can think of. Freyd's theorem implies that we cannot describe the homotopy category as a category of algebras for some algebraic theory $\mathcal{T}$, since any algebraic category is concrete.

Fun fact: Freyd's theorem essentially relies only on the general set-theoretic arguments and cardinal counting.

Non-counter-example: Whitehead's theorem states that if a map $f: X \to Y$ between pointed connected CW-complexes induces an isomorphism on all homotopy groups $\pi_i, \ i=1, 2,\dots$, then $f$ is a homotopy equivalence between $X$ and $Y$. Note that you still can't discriminate spaces looking just at the collection of homotopy groups: there can be $X$ and $Y$ such that $\pi_i(X) = \pi_i(Y)$ for all $i$, but this isomorphism is not induced by any actual map $F: X\to Y$ and the spaces are not homotopy equivalent. The simplest example is $\Bbb R \Bbb P^2 \times S^3$ and $S^2 \times \Bbb R \Bbb P ^3$. They both have a double covering by $S^2 \times S^3$ and have thus the same homotopy groups, but their cohomology is not isomorphic.

On second thought, I can't see why Freyd's theorem would imply actual non-discriminable spaces without any extra conditions on the invariant. Perhaps someone can fill this gap, but imho the non-discrimination of maps is bad enough.

Since this theorem states indiscriminability even for non-homotopy equivalent spaces, it in particular does so for non-homeomorphic spaces. This could be weaker than what your professor implied since we could in principle consider invariants of spaces which are not homotopy invariants. However, this requires some more specifics on what we would call "algebraic invariants" since e.g. the lattice of open subsets looks like a perfectly fine algebraic invariant to me, but it certainly discriminates spaces.

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    $\begingroup$ Freyd's result implies that no set of functors from the homotopy category of based spaces to the category of sets can discriminate between all morphisms. If you want to read this as "no matter how many such invariants ..." you should note that "many" means a set as opposed to a proper class. $\endgroup$ Mar 23, 2016 at 13:04
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    $\begingroup$ Another remark about Whitehead's Theorem: a map $X\to Y$ (of based connected CW complexes) is not necessarily nullhomotopic if it induces the null map on every $\pi_n$. For example, it may be that for every $n$ either $\pi_nX$ or $\pi_nY$ is trivial. One example is a torus mapping to a $2$-sphere. $\endgroup$ Mar 23, 2016 at 13:09
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    $\begingroup$ Another example of a map that is 0 in homotopy but not null is given by any cohomology operation in singular cohomology of non-zero degree. $\endgroup$ Mar 23, 2016 at 13:25
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    $\begingroup$ I don't understand how this counts as an answer to the question as posed. After all, the question was about homeomorphism invariants and the category $\mathsf{Top}$ does admit a faithful functor to $\mathsf{Set}$, namely the standard forgetful functor. How is non-existence of such functor for the homotopy category relevant? $\endgroup$ Mar 23, 2016 at 15:05
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    $\begingroup$ Let me clarify: if we interpret invariants as functors, then complete invariants correspond to conservative functors not faithful ones. I suspect that there isn't any conservative functor from either $\mathsf{Top}$ or $\mathsf{HoTop}$ to $\mathsf{Set}$, but this is not what Freyd's theorem is about. $\endgroup$ Mar 23, 2016 at 15:12

There are interesting jumps in the responses from the question to other questions, e.g. from homeomorphisms to homotopy equivalences, from spaces to pointed spaces, .... One reason for the homotopy study is that invariants up to homeomorphy might be uncountable, except for restricted classes of spaces, while for classes of "nice" spaces we can hope for a countable number of homotopy invariants.

In my own work I have been forced into considering "structured spaces" to get higher homotopy groupoids, and so colimit theorems, in higher homotopy; this pdf file on "A philosophy of modelling and computing homotopy types" is a relevant 2015 presentation. These notions of structured spaces, i.e. of topological data, extend to higher dimensions the idea of using many base points for the fundamental groupoid, and do lead to some not previously available precise nonabelian colimit calculations of homotopy types, and so of homotopy groups, of pointed spaces. It should not be forgotten that homotopy groups are but a pale shadow of pointed homotopy types.

Part of the philosophy is that in order to specify a space you need some kind of data, and that data will have some kind of structure. It could be wise to keep that structure in mind when constructing "invariants". It is for example standard to consider the cellular chain complex of a cell complex, and this is extended to crossed complexes and filtered spaces in the book Nonabelian Algebraic Topology. Grothendieck in Section 5 of his "Esquisse d'un Programme" argues that the notion of topological space is too far from the geometry, and wants more structure, such as a stratification.


Topological spaces can be infinitely varied. From topology point of view even something as deadpan as $\mathbb{R}$ has an enormous amount of structure.

What about the Zariski topology in algebraic geometry? They tend to be different than the topology you inherit from their embedding into ambient space.

  • $\begingroup$ I'm not sure why the downvotes- essentially this is the core of Anton Fetisov's answer. Namely, cardinality counting. $\endgroup$ Mar 23, 2016 at 15:33
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    $\begingroup$ @DanielMoskovich: I presume because the points here are still extremely imprecise — “infinitely varied”, “an enormous amount of structure”, “tend to be different” — and don’t even hint towards any precise statements. It takes a pretty generous reading to find the suggestion “cardinality counting” in this answer. $\endgroup$ Mar 24, 2016 at 12:57

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