Ranking topological invariants by "strength?"

Question

Is it possible to rank the common topological invariants (homology, cohomology, homotopy) according to their "strength?" By this I mean, can spaces have different homotopy groups yet the same homology and cohomology groups? Similarly, can spaces have different homology groups, yet the same homotopy and cohomology groups? And so on. I think the answer is that homology and cohomology groups determine each other, while homotopy groups can differ (making this a more fine/strong topological invariant). But I'm no expert by any means, so I'd really like to confirm this or be given a direction to further explore this. Thanks in advance.

Answer 1

To a connected topological space $X$, we can associate a) its sequence of homology groups $H_n(X;\mathbb{Z}),\,n\ge0$, b) its sequence of cohomology groups $H^n(X;\mathbb{Z}),\,n\ge0$ and c) its sequence of homotopy groups $\pi_n(X),\,n\ge0$. These are considered up to isomorphism.

The universal coefficient theorem tells us that a) determines b). This is the only relation between these invariants. Indeed, to see that b) does not determine a), we can take the Moore spaces $X=M(\mathbb{Q},1)$ and $Y=M(\mathbb{Q}^2,1)$ (see here). To see that c) does not determine b) (and hence c) does not determine a)), we can take $X=S^2\times\mathbb{RP}^3$ and $Y=S^3\times\mathbb{RP}^2$ (see here). To see that a) does not determine c) (and hence that b) does not determine c)), we can take $X=S^3$ and $Y$ the Poincaré homology sphere.

Answer 2

You might be interested in the category theoretic concept of a natural transformation.

Suppose you have two functors with the same domain category. I'm going to stick with a very basic example where the domain category is path connected topological spaces with a base point, and the two functors are the fundamental group $\pi_1$ and the first homology group $H_1$ (I'm not mentioning the morphisms, which you can surely guess).

We know that $\pi_1$ is "stronger" than $H_1$ in the sense that you ask, because isomorphic $\pi_1$ implies isomorphic $H_1$; this follows from the first Hurewicz theorem which says that $H_1(X)$ is the abelianization of $\pi_1(X,p)$.

But the first Hurewicz theorem actually says a little bit more: it says that abelianization is a natural transformation from the the functor $\pi_1$ to the functor $H_1(X)$. It not only converts the fundamental group into the first homology group, but if you are given a domain morphism, e.g. a continuous function $f : (X,p) \to (Y,q)$, then abelianization converts the $\pi_1$ homomorphism induced by $f$ into the $H_1$ homomorphism induced by $f$, and it does this in a fashion that respects the "category theoretic" structures, i.e. the compositions and the identities.

This concept of natural transformation is extremely useful, and among other things it fits right in with your concept of ranking invariants by strength: if there is a natural transformation from functor #1 to functor #2 then one can certainly say that #1 is stronger than #2.

And, by the way, $\pi_1$ is strictly stronger than $H_1$, because spaces can have isomorphic $H_1$'s but non-isomorphic $\pi_1$'s. Long before category theory Poincaré discovered a closed 3-manifold with trivial $H_1$ but nontrivial $\pi_1$, and this led him to fix his wrong conjecture and formulate his right conjecture.