A certain map involving Eilenberg-MacLane spaces is a well-defined homomorphism (towards a universal coefficient theorem)

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I'm working through Arkowitz's Introduction to Homotopy Theory for self-study. In the beginning of chapter 5 section 2, we are aiming for a proof of (a simplified version of) the Universal Coefficient Theorem for (homotopical) cohomology (where $H^n(X;G)\equiv[X,K(G,n)]$ with $K(G,n)$ is an Eilenberg-MacLane space). Along the way, we're asked to prove the following.

Question

For a space $X$ and an abelian group $G$, we define a homomorphism $\alpha = \alpha_G : H^n(X) \otimes G \rightarrow H^n(X;G)$ as follows. Given $[f] \in H^n(X) = [X, K(\mathbb{Z},n)]$ and $\gamma \in G$, then $\gamma$ induces a homomorphism $\theta_\gamma: \mathbb{Z} \rightarrow G$ defined by $\theta_\gamma(1)=\gamma$. By earlier lemmas, we know that $\theta_{\gamma}$ determines a unique homotopy class $[\phi_\gamma] \in [K(\mathbb{Z},n),K(G,n)]$ given by $\phi_\gamma* = \theta_\gamma:\pi_n(K(\mathbb{Z},n))\rightarrow\pi_n(K(G,n))$. We set $\alpha([f]\otimes \gamma)=[\phi_\gamma f]$. Show $\alpha$ is a well-defined homomorphism by showing that

Part 1: $\phi_\gamma(f+f^\prime)\simeq\phi_\gamma(f) + \phi_\gamma(f^\prime)$

Part 2: $\phi_{\gamma + \gamma^\prime} (f)\simeq\phi_{\gamma}(f) + \phi_{\gamma^\prime}(f) $

Thoughts

One thing about this book is that it forces the reader, at key moments, to go back and review many of the earlier concepts/proofs to make sure that they are solid before proceeding. I believe I have a straightforward solution to Part 2, but remain stuck on Part 1.

For Part 1, maybe I should begin by replacing $K(\mathbb{Z},n)$ with $\Omega K(\mathbb{Z},n+1)$ and replace $K(G,n)$ with $\Omega K(G,n+1)$, just so that the addition operation is apparent. From here some ideas occur, but none have panned out so far.

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To close this out, after reading the comment and some further discussion/review, I'll sketch a proof.

We require a Lemma (5.2.1) from the text

Lemma

If $G$ and $H$ are abelian groups and $f,g: K(G,n)\rightarrow K(H,n)$ are two maps, then $f_*=g_*:\pi_n(K(G,n))=G\rightarrow\pi_n(K(H,n))=H, n\geq1$, if and only if $f\simeq g$.

Proof of Part 1

We begin by observing that $$\phi_\gamma \circ(f+f^\prime)\equiv \phi_\gamma \circ\mu_{K(\mathbb{Z},n)}\circ (f\times f^\prime) \circ \Delta$$ where $\mu_{K(\mathbb{Z},n)}$ is the multiplication in $K(\mathbb{Z},n)$ and $\Delta$ is the diagonal map. Also

$$\phi_\gamma f+\phi_\gamma f^\prime\equiv \mu_{K(G,n)} \circ (\phi_\gamma f\times \phi_\gamma f^\prime) \circ \Delta= \mu_{K(G,n)} \circ (\phi_\gamma \times \phi_\gamma) \circ (f\times f^\prime) \circ \Delta.$$

So it will suffice to show that

$$\phi_\gamma \circ\mu_{K(\mathbb{Z},n)} \simeq \mu_{K(G,n)} \circ (\phi_\gamma \times \phi_\gamma): K(\mathbb{Z\times Z},n)\rightarrow K(G,n). $$

We note that $K(\mathbb{Z\times Z},n) \simeq K(\mathbb{Z},n) \times K(\mathbb{Z},n)$. For $(z, z^\prime) \in \pi_n(K(\mathbb{Z},n)) \times \pi_n(K(\mathbb{Z},n))$:

$$(\phi_\gamma \circ\mu_{K(Z,n)})_* [(z,z^\prime)]= \theta_\gamma [z + z^\prime] = \theta_\gamma [z] + \theta_\gamma [z^\prime] = (\mu_{K(G,n)} \circ (\phi_\gamma \times \phi_\gamma))_* [(z,z^\prime)] $$ and we now apply the lemma.