Can we classify all finite CW complexes $X$ such that for each $i$ there is some isomorphism $\pi_i(X) \rightarrow H_i(X)$? Note that it is not hard to classify all complexes for which each isomorphism is given by the Hurewicz map.
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asked Oct 8, 2010 at 14:39
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1$\begingroup$ @Manuel, do you have a particular motivation for this problem? The answer is going to be some small class of CW-complexes but I'm not seeing a motivation for the question. In that regard the title is a little misleading because homotopy groups form a type of graded lie algebra and homology groups don't. Moreover, you're asking for a dimension-wise isomorphism of groups, not an equality of groups. $\endgroup$– Ryan BudneyCommented Oct 8, 2010 at 15:07
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$\begingroup$ Would we have to know, a priori, all of the homotopy groups of a space to answer this question? $\endgroup$– JoshCommented Oct 8, 2010 at 15:11
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4$\begingroup$ The "right" viewpoint on invariants like $H_i$ and $\pi_i$ is that they are functorial invariants. It is somehow "wrong" to not think of them functorially. The Whitehead theorem gives some justification for this philosophy. $\endgroup$– Kevin H. LinCommented Oct 8, 2010 at 15:19
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$\begingroup$ Yes, certainly the question sounds like it lacks of motivation, since crazy maps between homotopy and cohomology and not particularly useful. I was just trying to think what is happening geometrically when you have these "non natural" isomorphisms. $\endgroup$– Manuel RiveraCommented Oct 8, 2010 at 15:29
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3$\begingroup$ @ Manuel - Just to give slightly different example : A space $X$ and the space $\prod K(\pi_i(X),i)$ have isomorphic homotopy groups. Unless you have a map from one to the other realizing these isomorphisms, there is no geometric meaning to these isomorphisms. The same goes for homology too - given a suitable $X$ you can cook up a product of Moore spaces which have homology isomorphic to $X$. But until you have a map one way or another that realizes these isomorphisms, it is not veru helpful. $\endgroup$– Somnath BasuCommented Oct 8, 2010 at 16:21
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I also don't see a motivation for this, but anyway: if $i=0$ is allowed, there is not a single example ($\pi_0$ is always finite, $H_0$ never. For the empty space, $\pi_0$ is empty and $H_0$ is not). Apart from $i=0$, the only connected finite CW-complexes with only finitely many (edit: abelian) homotopy groups are tori. (J.-P. Serre, Cohomologie modulo 2 des complexes d’Eilenberg-MacLane, Comment. Math. Helv. 27 (1953), 198-232.)
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1$\begingroup$ Not just tori have finitely many non-trivial homotopy groups (e.g., surfaces of positive genus, etc.). Do you mean these are the only ones with abelian $\pi_1$? $\endgroup$ Commented Oct 8, 2010 at 22:08