For which spaces is the natural map $\pi_n(X) \to H_n(X)$ an isomorphism in all degrees?

Solution 1:

This answer was co-written with Michael Andrews at UCLA; thanks for his help on this.

This is possible if and only if $X$ is a product of factors of the form $K(\Bbb Q, 2n+1)$ ($n \geq 1$), $K(\Bbb Z_{p^\infty}, 2n)$ (as $p$ ranges over odd primes and for $n \geq 1$) and $K(\Bbb Z[1/S], 1)$ ($S$ a set of primes), with at most one of each factor (that is, $K(\Bbb Q, 3)$ and $K(\Bbb Q,5)$ are not both allowed, nor are two copies of $K(\Bbb Q,3)$, but $K(\Bbb Z_{3^\infty}, 2) \times K(\Bbb Z_{5^\infty}, 2)$ is OK), a factor of the first type only allowed if there are no factors of the third type, and a factor of the second type at the prime $p$ only allowed if $p$ is in the set of primes $S$. (Note that the statement in the question that it cannot be $K(G,2n)$ is NOT true - that's only true when $G$ is the additive group of a ring, which the Prufer groups are not!)

As noted in the comments, the infinite symmetric product $\text{SP}^\infty(X)$ has homotopy groups naturally isomorphic to the homology groups of $X$, and the Hurewicz map is induced by the map $X = \text{SP}^1(X) \to \text{SP}^\infty(X)$. So we're asking precisely when this map is a (weak) homotopy equivalence. $\text{SP}^\infty(X) = \prod_{i=1}^\infty K(H_i(X;\Bbb Z), i)$, so necessarily $X$ is a product of Eilenberg-MacLane spaces.

If $X \times Y$ has the desired property, so do the factors: the image of the Hurewicz map to $X \times Y$ lies in $H_k(X) \oplus H_k(Y)$ (with each factor of $\pi_k(X) \oplus \pi_k(Y)$ going to the obvious place in homology). So we should start by figuring out which Eilenberg MacLane spaces are also Moore spaces.

Let's suppose $n>1$ for now. We'll return later to the $n=1$ case.

First, $G$ is divisible. For suppose the multiplication-by-p map was not surjective; then $G/pG$ is a $\Bbb F_p$-vector space, so we can get a map $G \to \Bbb F_p$. This induces a map $K(G,n) \to K(\Bbb F_p, n)$, which one verifies is nonzero on $H^n( -; \Bbb F_p)$. Let $\iota \in H^n(K(\Bbb F_p, n); \Bbb F_p)$ be the fundamental class. Then we know that, for instance, the Steenrod power $P^1\iota$ is nonzero if $p > 2$, and thus by naturality, since we know there is an element $\alpha \in K(G,n)$ that pulls back to $\iota$, we see that $P^1\alpha$ must be nonzero. For $p=2$, $\text{Sq}^2\iota$ suffices to perform the same argument. Thus $K(G,n)$ has cohomology in degree $n+2$ or higher, contradicting the Moore hypothesis.

Therefore, $G = \Bbb Q^n \oplus_p \Bbb Z_{p^\infty}^{n_p}$. Let's calculate the homology of the factors. $H_*(K(\Bbb Q, 2n+1)) = \pi_*(K(\Bbb Q, 2n+1))$, as desired; $H^*K(\Bbb Q,2n)$ is a polynomial algebra on a degree $2n$ generator. Now we calculate for $G = \Bbb Z_{p^\infty}$. We have that $K(G,n) = \text{colim}_{k} K(\Bbb Z/p^k, n)$, and homology commutes with directed colimits. $K(\Bbb Z/p^k, n)$ has integral homology that is a finite $p$-group for all $p$; the homology of $K(G,n)$, then, has to be a colimit of finite $p$-groups, thus being a direct sum of copies of $\Bbb Z/p^k$ and $\Bbb Z_{p^\infty}$. In any case, $\text{Tor}(\Bbb Z/p^k, \Bbb F_p)$ is $\Bbb F_p$, as is $\text{Tor}(\Bbb Z/p^\infty, \Bbb F_p)$ (because $\text{Tor}$ commutes with directed colimits, and the induced map on the tor groups of $\Bbb Z/p^k \to \Bbb Z/p^{k+1}$ is the identity.) Thus $K(G,n)$ is a Moore space iff its $\Bbb F_p$ homology is zero outside of degree $0, n+1$ and is $\Bbb F_p$ in those degrees.

So let's prove that this is true when $p$ and $n+1$ are odd. Again, homology commutes with colimits, so we need to compute the induced map on $\Bbb F_p$ homology of $K(\Bbb Z/p^k, n) \to K(\Bbb Z/p^{k+1}, n)$. The easiest way to do this is to actually work with $\Bbb F_p$ cohomology (since the universal coefficient theorem with field coefficients says that the induced map on $\Bbb F_p$ cohomology is just the transpose of the induced map on $\Bbb F_p$ homology). Then the $\Bbb F_p$ cohomology of these spaces is a free algebra on classes given by cohomology operations applied to the fundamental class corresponding to the "reduction mod $p$" homomorphism $\Bbb Z/p^k \to \Bbb Z/p$. This fundamental class maps to zero (it maps to the composite $\Bbb Z/p^k \to \Bbb Z/p^{k+1} \to \Bbb Z/p$). The cohomology operations involved are all either 1) the Bockstein operation corresponding to $\Bbb Z/p \to \Bbb Z/p^{k+1} \to \Bbb Z/p^k$ or 2) that operation, after first applying one of the standard mod $p$ operations. (See this paper.) Call the first $Q_k$. Then the induced map on cohomology must vanish on $P\iota_n$, since mod $p$ cohomology operations are natural, and the induced map kills $\iota_n$. The only one which can fail to vanish is the map on $Q_{k+1}\iota_n$, which maps to $Q_k \iota_n$ (it's not natural, since it depends on the $K(\Bbb Z/p^k, n)$ we're living in!) When $n+1$ and $p$ are odd, the free subalgebra generated by this element is the exterior algebra and thus gives no new elements; when $p$ or $n+1$ is even, it's a polynomial algebra, thus giving many new nonvanishing elements. Thus when $p$ is odd and $n+1$ is even in every degree but $H^{n+1}$, the maps are all zero; the same is true in homology.

Since the maps in the colimit are all either zero (in degrees other than $n+1$) or an isomorphism $\Bbb F_p \to \Bbb F_p$ (in degree $n+1$), the colimits are zero in every degree other than $k+1$, where the final result is $\Bbb F_p$, as desired. It is otherwise false, since we get a whole polynomial algebra from the element surviving in $H^{n+1}$.

Now for the $n=1$ case. Obviously $G$ must be abelian. One could generalize to the case where we only demand the question in degree greater than 1, but I don't care to. It is apparently an open problem to prove that groups with homological dimension $1$ are locally free (that is, every finitely generated subgroup is free); I do not know if the abelian case is known, but if it is, this would be the same as saying that an abelian group has homological dimension 1 iff every finitely generated subgroup is trivial or $\Bbb Z$. This would then imply that the group is a subgroup of $\Bbb Q$, which are up to isomorphism the same as localizations of $\Bbb Z$ at some set of primes. I'm not going to look much more into this.

Now let's think about products when $n \geq 2$; we'll deal with $n=1$ later. Write $Q_n = K(\Bbb Q, 2n+1)$, and $X_{p,n} = K(\Bbb Z_{p^\infty}, 2n+1)$. Then $X_{p,n} \times X_{q,m}$ is also one of our spaces when $p \neq q$, since the tensor product of Prufer groups is automatically zero, and $\text{Tor}(\Bbb Z_{p^\infty}, \Bbb Z_{q^\infty}) = 0$. On the other hand, $\text{Tor}(\Bbb Z_{p^\infty}, \Bbb Z_{p^\infty}) = \Bbb Z_{p^\infty}$, so the Kunneth theorem implies that a product $X_{p,n} \times X_{p,m}$ is not one of our spaces (it's got extra homology in degree $n+m-1$.) We can multiply in at most one copy of $Q_n$; two violates Kunneth (we get a copy of $\Bbb Q \otimes \Bbb Q = \Bbb Q$ in degree $m+n$), but $\text{Tor}(\Bbb Z_{p^\infty}, \Bbb Q) = 0$, and $\Bbb Q \otimes \Bbb Z_{p^\infty} = 0$.

So a simply connected example is a product of at most one $Q_n$ and one (for each $p$) of the $X_{p,n}$. A non-simply connected example, assuming the conjecture stated above, can only exist when there is no $Q_n$ factor (or we'd get too much homology from Kunneth) - and then, if the fundamental group is $\Bbb Z[1/S]$, where $S$ is a set of primes, then no $X_{p,n}$ factor can appear where $p$ is not in $S$ - again, tensor products cause problems. (In particular, if the fundamental group is $\Bbb Z$, we can only have the circle $S^1$.)