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The following theorem of Hurewicz holds (let $\cdot$ be the one-point space and $n\!\geq\!2$):

If $\pi_i(X)\cong\pi_i(\cdot)$ for $i\!<\!n$, then $H_i(X)\cong H_i(\cdot)$ for $i\!<\!n$ and $\pi_n(X)\cong H_n(X)$.

Does the following hold (I suspect not):

(1) If $\pi_i(X)\cong\pi_i(Y)$ for $i\!<\!n$, then $H_i(X)\cong H_i(Y)$ for $i\!<\!n$, and $\pi_n(X)\cong H_n(X)$?

Does the following hold:

(2) If $\pi_i(X)\cong\pi_i(Y)$ for all $i$, then $H_i(X)\cong H_i(Y)$ for all $i$?

Does the following hold:

(3) If $H_i(X)\cong H_i(Y)$ for all $i$, then $\pi_i(X)\cong \pi_i(Y)$ for all $i$?

If yes, how can I prove them, preferrably using the Hurewicz theorem above. If not, are some additional mild hypotheses sufficient to make it work?

Do (2') and (3') hold, in which $H_i$ has been replaced by $H^i$?

Grigory M
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Leo
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2 Answers2

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(2) is true if the isomorphisms on $\pi_n$ are induced by a map $f\colon X\to Y$. This follows from Whitehead's theorem. But in general the answer is no. Wikipedia's article on the Whitehead theorem gives the example of $X=S^2\times \mathbb{RP}^3$ and $Y=\mathbb{RP}^2\times S^3$. They have isomorphic homotopy groups, since they have the same fundamental group $\mathbb Z_2$, and have the same universal covers. But they have different homology groups, which you can see by using the Künneth formula.

Cheerful Parsnip
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  • Can Hurewicz be ever used for non-simply-connected spaces? I'm confused about the usefulness of the theorem. It's supposed to be the cornerstone of AlgTop. But it seems we can apply it in a very narrow set of examples. Perhaps this should be my initial question... – Leo Jun 27 '13 at 11:38
  • In the non-simply connected case the Hurewicz theorem turns into the statement that $H_1$ is the abelianization of $\pi_1$. – Cheerful Parsnip Jun 27 '13 at 13:16
  • And that's it? No correspondence between higher $\pi_k$ and $H_k$? – Leo Jun 27 '13 at 16:59
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    Well there are some things you can say. For example if $\pi_1\neq 0$ and $\pi_{\geq 2}=0$ you get a $K(\pi_1,1)$ space, whose homology is uniquely determined. (That's one way to define the homology of a group.) – Cheerful Parsnip Jun 27 '13 at 19:06
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(1) does not hold. For the last part ($\pi_n(X) \simeq H_n(X)$), simply take $X=Y$ and find a counterexample to Hurewicz's theorem if you don't assume $\pi_k(X) = 1 \forall k < n$. For the first part, take the following counterexample (with a big enough $n$ st. $H_n(X) \not\simeq H_n(Y)$):

(2) does not hold either. For example, $X = S^2 \times \mathbb R P^3$ and $Y = S^3 \times \mathbb R P^2$ have the same homotopy groups, but not the same homology (see Wikipedia's article and Grumpy Parsnip's answer). The additional condition needed is that the isomorphisms $\pi_*(X) \simeq \pi_*(Y)$ must be induced by a map $f : X \to Y$.

(3) does not hold either. $S^2 \times S^4$ and $\mathbb C P^3$ have the same homology groups ($\mathbb Z$ in degrees 0,2,4,6 and zero elsewhere), but different homotopy groups ($\pi_2(\mathbb C P^3) = \mathbb Z$ but $\pi_2(S^2 \times S^2) = \mathbb Z^2$).

For (1) and (3) I don't know of nice additional conditions that would make the theorems true. Note that the counterexamples to (2) and (3) also hold in cohomology (use the universal coefficient theorem).

Najib Idrissi
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