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In the same vein of this MO question, one can ask:

If two spaces $X$, $Y$ have isomorphic generalized cohomology rings $\mathrm{h}^{\bullet}(X)\cong \mathrm{h}^{\bullet}(Y)$ for every multiplicative generalized cohomology theory $\mathrm{h}^{\bullet}$, do they have to be homotopically equivalent?

What if we also take cohomology operations into account?

(It's just a spontaneous question that popped in my mind, as a non expert in topology)

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Oh, I just noticed there's a conflict between the question in the title and the question in body of the message... Maybe I should edit. –  Qfwfq Mar 2 '11 at 17:30
(Edited: I kept the title and changed the body, and also likewise edited the answers for the benefit of the readers) –  Qfwfq Mar 2 '11 at 17:37

3 Answers 3

up vote 13 down vote accepted

No, they don't have to be homotopically equivalent. In fact:

  • There is a map of CW-complexes $X \to Y$ which is an isomorphism on (co)homology, full stop, for every generalized (co)homology theory $h$, multiplicative or not.
  • This is, in fact, equivalent to the map $X \to Y$ being an isomorphism on integral homology.
  • Systematic examples of such maps are given by plus constructions.
  • However, if the spaces involved are connected and simply connected, such a map must be a homotopy equivalence by the homology Whitehead theorem.

This leads to a more subtle question, which is:

  • Are there two simply-connected spaces whose cohomology rings are isomorphic for all generalized cohomology theories $h$, but which are not homotopy equivalent?

Then the answer is yes. The spaces $S^3 \vee S^5$ and $S^1 \wedge {\mathbb{CP}}^2$ have the same cohomology rings for all generalized cohomology theories, but they're not homotopy equivalent. This is the suspension of Neil Strickland's example in the other question, which has the effect of killing all the multiplication in the cohomology ring.

EDIT: In the new version, where cohomology operations are allowed, the question becomes more difficult because you're moving in the right direction: attaching more data like cohomology operations (the next to allow would be secondary operations). I think this example covers that case: you have two spaces, $S^3 \vee S^3 \vee S^8$ and another one formed by some Whitehead-product cell attachment, whose cohomology rings both have trivial multiplication. The cohomology operations on the Whitehead-product space are all trivial because the 8-cell is attached to both copies of $S^3$, but not to either one individually: specifically, if you collapse down either copy of $S^3$ to a point you're left with something homotopy equivalent to $S^3 \vee S^8$.

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Thank you. Interesting! (I accept this answer, among the others, just because this is more expanded but the idea was present also in the others) –  Qfwfq Mar 2 '11 at 17:28
I'd like to point out a special case of your plus construction example: how about $M^3 =$ Poincare's homology sphere? This has a degree one map $M^3 \to S^3$ which is a multiplicative cohomology isomorphism in any theory. –  John Klein Mar 2 '11 at 19:22
Let $X$ be a noncontractible space such that $\Sigma X \simeq *$. Then every cohomology theory will see $X$, incorrectly, as contractible. There are such spaces. –  Jeff Strom Mar 2 '11 at 20:59
$S^3\vee S^5$ and $\Sigma\mathbb{CP}^2$ do not agree on all cohomology theories--they are not stably equivalent. Specifically, any spectrum on which the Hopf map acts nontrivially (eg, the sphere spectrum) will give a cohomology theory that distinguishes them. –  Eric Wofsey Mar 2 '11 at 22:50
@Eric: Yes, you're correct - that's a ridiculous mistake on my part. I won't edit it since I don't want to bump it, but I suppose the second example will have to suffice - or any 2-cell complex formed by attaching a sphere where the attaching map becomes stably trivial. –  Tyler Lawson Mar 3 '11 at 2:06

If you restrict to CW-complexes, something strong is true: if a map $f:X \to Y$ induces an iso on integral homology, then its suspension $SX\to SY$ a homotopy equivalence, and thus the map f is iso in every generalized homology/cohomology theory.
There are many examples of such f which are not homotopy equivalences.

If we require the spaces to be simply connected, the question becomes more interesting.

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No – Knot complements provide a counterexample. The cohomology is that of $S^1$ and hence, by the Atiyah-Hirzebruch spectral sequence, the same holds for any generalized cohomology theory. Also there cannot be any interesting products. On the other hand, knot complements determine knot types by the Gordon-Luecke theorem, and there are certainly knots different from the unknot.

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Charles and Tyler were faster. –  Tilman Mar 2 '11 at 17:17

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