*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.

precisedefinition 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. $\endgroup$ – Francesco Polizzi Mar 23 '16 at 9:0513more comments