Is every simplicial map $\Phi:K(A, n) \to K(A', n)$ a simplicial homomorphism of groups?

Question

I have posted a few questions on MSE, most notably this one, which revolve around the same issue and have received no answers, so I decided to ask the same here.

In the following, $K(A, n)$ is the minimal Eilenberg-MacLane Kan complex given by $$K(A, n)_q=\{\text{normalized $n$-cocycles } \Delta^q \to A\}.$$ This is constructed via the Dold-Kan correspondence, which applies to show that $\mathsf{sAb}(K(A, n), K(A', n)) \cong \mathsf{Ab}(A, A')$. In light of proving the existence of an isomorphism $\mathsf{sSet}(K(A, n), K(A', n))\cong \mathsf{Ab}(A, A')$, the following issue arises.

Is every simplicial map $\Phi:K(A, n) \to K(A', n)$ a simplicial homomorphism of groups? May in his book Simplicial objects in Algebraic Topology, Lemma 25.1, claims that the answer is positive and I know no other reference for this result. The proof is by induction: in degrees $q<n$, $K(A, n)_q=\{\ast\}=K(A', n)_q$ and there is nothing to prove. By minimality, $K(A, n)_n=\pi_n(K(A, n))$ and $\pi_n(\Phi)=\Phi_n$, hence $\Phi_n$ is a homomorphism. Now come the problems: suppose for $q\ge n+1$ that we have proven that $\Phi_{q-1}$ is homomorphism. Consider $\Phi_q(x+y)$ and $\Phi_q(x)+ \Phi_q(y)$. These two elements of $K(A', n)$ have the same boundary by the induction hypothesis. May concludes that they are homotopic. Why this? We know that $\pi_q(K(A', n)\cong 0$, but $\Phi_q(x+y)$ and $\Phi_q(x)+ \Phi_q(y)$ need not determine elements of $\pi_q(Y)$, since there is no reason why their boundary should be constant at the base-point. If we accept that they are homotopic relative their boundary, then they are equal by minimality and the argument is concluded. I don't want to sound skeptical about May's proof, but I am not convinced, and I even don't know if the result itself is true.

Why I doubt this is true:

• It is well-known that $[K(A, n), K(A', n)]\cong H^n(K(A, n), A')\cong \mathsf{Ab}(A, A')$. If the isomorphism above holds true, then we get an isomorphism $[K(A, n), K(A', n)] \cong \mathsf{sSet}(K(A, n), K(A', n))$. (And this groups can be finite.) Using the explicit form of the involved isomorphisms, this essentially says that two homotopic maps $K(A, n) \to K(A', n)$ are equal, which I highly doubt to be true, even using minimality of the codomain.
• The only other place I know in literature where something similar is treated is a book (in German) by Lamotke, Semisimpliziale Algebraische Topologie. In Theorem (Satz) VIII.3.11, he proves that every simplicial map $K(A, n) \to K(A', n)$ is homotopic to a homomorphism and every two homotopic homomorphisms are equal. Although this does not rule out the possibility that every simplicial map $K(A, n) \to K(A', n)$ be a homomorphism on the nose, if this is true, why not saying it?

So, my question boils down to: is this fact true, and if yes, why?

[Edit: I have now removed the question on MSE.]

Answer 1

Denote by $$\def\U{{\sf U}}\def\sSet{{\sf sSet}}\def\sAb{{\sf sAb}}\def\Ch{{\sf Ch}}\def\N{{\sf N}}\U\colon \sAb→\sSet$$ the forgetful functor, which is the right adjoint of $$\def\Z{{\bf Z}}\Z\colon\sSet→\sAb$$ and by $$Γ\colon \Ch→\sAb$$ the Dold–Kan functor given by the right adjoint of the normalized chains functor $$\N\colon\sAb→\Ch.$$ Thus $$\def\K{{\sf K}}\K(A,n)=\U Γ(A[n]).$$

We have a chain of isomorphisms of sets $$\sSet(\K(A,n),\K(A',n))≅\sSet(\U Γ A[n], \U Γ A'[n])≅\Ch(\N\Z\U Γ A[n],A'[n])$$ (by adjunctions $\Z⊣\U$, $\N⊣Γ$), $$\def\Hom{\mathop{\sf Hom}}\def\H{{\sf H}}\Ch(\N\Z\U Γ A[n],A'[n])≅\Hom(\H_n(\N\Z\U Γ A[n]),A')$$ (since both chain complexes vanish below degree $n$), and if $n>0$, we have $$\Hom(\H_n(\N\Z\U Γ A[n]),A')≅\Hom(A,A')$$ (by the Hurewicz theorem). This proves the desired isomorphism $$\sSet(\K(A,n),\K(A',n))≅\Hom(A,A').$$ (If $n=0$, then $\Hom$ refers to maps of sets, not groups.) The same argument also proves that $$\sSet(\K(A,m),\K(A',n))≅0$$ (i.e., the only simplicial map factors through the basepoint) if $m>n>0$. If $m<n$, then the Hurewicz theorem is not applicable, and in this case we have nontrivial cohomology operations.