Is there an analogue of Eilenberg-Maclane spaces for homology?

Question

Let $G$ be a group and $n$ a positive integer. A connected topological space $Y$ is called an Eilenberg–MacLane space of type $K(G, n)$, if $\pi_n(Y) \cong G$ and all other homotopy groups of $Y$ are trivial; as such as spaces are unique up to weak homotopy equivalence, we often denote such a space by $K(G, n)$.

If $G$ is an abelian group and $X$ is a CW complex, then $$[X, K(G, n)] \cong H^n(X; G);$$ where the cohomology is singular cohomology.

Does there exist a family of topological spaces $J(G, n)$ which satisfy a similar relationship for singular homology? More precisely,

Let $G$ be an abelian group and $n$ a positive integer. Does there exist a connected topological space $J(G, n)$ such that, for any CW complex $X$, $$[J(G, n), X] \cong H_n(X; G)?$$ If so, is it unique up to homotopy equivalence? Can $J(G, n)$ be characterised in other ways (as is the case for Eilenberg-MacLane spaces)?

Answer 1

If you want to replace homotopy classes of maps out with homotopy classes of maps in, what you get is homotopy, not homology. This is part of the yoga of Eckmann-Hilton duality. Because it's a "maps-in" construction, homotopy behaves well with respect to homotopy limits (e.g. the long exact sequence of a fibration). Both homology and cohomology, on the other hand, behave well with respect to homotopy colimits (e.g. the Mayer-Vietoris sequence).

The categorical way to think about homology is to think of it as coming from a higher analogue of taking the free abelian group on a set (rather than taking either functions in or out of the set). There are a number of ways to make this precise, such as the Dold-Kan theorem and the Dold-Thom theorem, as well as the more abstract approach involving spectra, which are the higher analogues of abelian groups.

You can think of taking the free abelian group as a "tensor product" operation $\mathbb{Z} \otimes X$, which among other things gives you the right expectations as to its behavior with respect to limits and colimits (it preserves colimits in the $X$ variable). The way this generalizes to spectra is that if $E$ is a spectrum, then $E$-homology is the homotopy groups of the "derived tensor product" $E \otimes X$ (formally, the smash product of $E$ with the suspension spectrum $\Sigma^{\infty}_{+} X$, which you should think of as the free spectrum on $X$). If $E$ is a ring spectrum, this can be thought of as the free $E$-module spectrum on $X$.

It might help to organize everything into the following table, which also includes the analogous operations in homological algebra:

Homotopy   | Ext(A, -) | Maps in  | Covariant     | Preserves limits 
Homology   | Tor(A, -) | Tensor   | Covariant     | Preserves colimits
Cohomology | Ext(-, A) | Maps out | Contravariant | Sends colimits to limits

(I should clarify that the statements in the last column aren't literally true as written; first, limits and colimits should be replaced by homotopy limits and colimits, and second, depending on whether you take homotopy groups or not "preserves" should be replaced by the existence of a nice spectral sequence.)

Answer 2

There can't be such a $J$ for any $n>0$.

For if that were the case, then $$H_n(X \times Y) \cong [J(G,n), X \times Y] \cong [J(G,n), X] \times [J(G,n), Y] \cong H_n(X) \times H_n(Y).$$

But this is obviously not true; if $Y = \{0, 1\}$ this would imply $H_n(X) \oplus H_n(X) \cong H_n(X \sqcup X) \cong H_n(X)$ for $n>0$.

If you want a connected example, one can show $H_n(X \times S^1) \cong H_n(X) \oplus H_{n-1}(X)$. This is usually not $H_n(X) \oplus H_{n}(S^1)$; you can pick $X$ so that this is not true for all $n>1$.