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It is well-known that for two (discrete) abelian groups $G$ and $H$, the set $[K(G,n),K(H,n)]_*$ of based homotopy classes of maps between the corresponding Eilenberg-MacLane spaces is in canonical bijection with ${\rm Hom}_{\rm Ab}(G,H)$, the set of group homomorphisms from $G$ to $H$. The usual textbook proof uses that $[X,K(H,n)]_* \cong H^n(X;H)$ and then uses the Universal Coefficients Theorem and Hurewicz to see $H^n(K(G,n);H) \cong {\rm Hom}_{\rm Ab}(G,H)$.

Question: Do we have a similar result if we replace homotopy classes of based maps $K(G,n) \to K(H,n)$ with homotopy classes of group homomorphisms $K(G,n) \to K(H,n)$?

The model I want to take for $K(G,n)$ is the $G$-linearization $G[S^n]$ of $S^n$, introduced by McCord as $B(G,S^n)$. Its elements are formal sums $\sum_{i=1}^ng_i x_i$ with $g_i \in G$ and $x_i \in S^n$, with the expected identifications and topology. It is a topological abelian group by adding formal sums.

That this topological group is indeed a model for $K(G,n)$ follows from a theorem of McCord (1969), which gives us a fiber sequence $G[S^{n-1}] \to G[D^n] \to G[S^n]$ induced by the cofiber sequence $S^{n-1} \hookrightarrow D^n \to D^n/S^{n-1} \cong S^n$. We conclude that there is a weak equivalence $G[S^{n-1}] \to \Omega(G[S^n])$ and by induction $G \simeq \Omega^n(G[S^n])$, showing that $G[S^n]$ is a $K(G,n)$ as desired.

Now a more precise formulation of my question is:

Question: Do we have a bijection $\pi_0({\rm Hom}_{\rm TopAb}(G[S^n],H[S^n])) \cong {\rm Hom}_{\rm Ab}(G,H)$?

There are obvious candidates for the maps in both directions. From left to right, we send $f: G[S^n] \to H[S^n]$ to $\pi_n(f): G \to H$. Conversely, a group homomorphism $\phi: G \to H$ is sent to $\phi[S^n]: G[S^n] \to H[S^n]$. Clearly $\pi_n(\phi[S^n]) = \phi$ for all $\phi: G \to H$. However, it is not so clear whether every continuous group homomorphism $f: G[S^n] \to H[S^n]$ is homotopic (through group homomorphisms) to one of the form $\phi[S^n]$.

Possible approach: Since we have an adjunction ${\rm Hom}_{\rm TopAb}(G[S^n],H[S^n]) \cong {\rm Hom}_{\rm TopAb}(G,\Omega^n(H[S^n]))$, where $\Omega^n(H[S^n])$ has the pointwise abelian group structure, it would suffice to prove that ${\rm Hom}_{\rm TopAb}(G,-)$ preserves the weak equivalence $H \xrightarrow{\simeq} \Omega^n(H[S^n])$. My idea would be to do this in the following steps:

  1. This is clear for $G = \mathbb{Z}$ since ${\rm Hom}(\mathbb{Z},A) \cong A$.
  2. Similarly, for $G = \mathbb{Z}/k$, we can use that ${\rm Hom}(\mathbb{Z}/k,A) \cong {\rm Tor}_k(A)$ and apply the above result to the abelian group ${\rm Tor}_k(H)$ instead of $H$.
  3. This means the result is true for any finitely generated $G$.
  4. By writing $G$ as colimit over its f.g. subgroups, it follows for any $G$.

However, I have not been able to find this result in the literature and thus I'm fearing I'm overlooking something. Can someone tell me whether this approach works? Also a reference to a similar result in the literature would be nice.

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    $\begingroup$ The model category of topological groups is Quillen equivalent to simplicial groups (via the usual nerve-realization adjunction), and the latter is Quillen equivalent to chain complexes via the Dold–Kan correspondence. Under this chain of equivalences, G[S^n] becomes weakly equivalent to G[n]. Thus, the desired set of maps can be computed as the derived hom in chain complexes from G[n] to H[n], which is precisely Hom(G,H). $\endgroup$ Jan 2, 2020 at 18:38
  • $\begingroup$ @DmitriPavlov Thanks for the confirmative answer to the question! I have to say I'm not so familiar with the derived set of maps, so I don't understand why this gives precisely the set of path components of Hom(G[S^n],H[S^n]) in the category TopAb. Could you comment on this? $\endgroup$ Jan 2, 2020 at 19:25
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    $\begingroup$ Quillen equivalences preserve derived mapping spaces. In particular, they preserve sets of connected components of derived mapping spaces. The mapping space in the statement happens to be derived, so we might as well compute it in chain complexes. $\endgroup$ Jan 2, 2020 at 21:17

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