$C_n(X;G)$ is naturally isomorphic to $C_n(X) \otimes G$

Question

Let $X$ be a space, and let $G$ be a fixed group. What does "$C_n(X;G)$ is naturally isomorphic to $C_n(X) \otimes G$" means?

I know that these two groups are isomorphic, since the following hold:

1. $C_n(X;G)$ is a direct sum of $G$'s,

2. $C_n(X)$ is a direct sum of $\Bbb Z$'s,

3. $\Bbb Z \otimes G \cong G$,

4. $\otimes$ commutes with $\oplus$.

But what is naturally isomorphic? Is this relevant with the notion "natural transformation"? Should I have to interpret as follows?

Given a continuous map $f:X \to Y$, the induced map $f_{\#} : C_n(X;G) \to C_n(Y;G) $ corresponds to the map $f_{\#} \otimes1_G:C_n(X)\otimes G \to C_n(Y)\otimes G$, under the above isomorphism $C_n(-;G)\cong C_n(-)\otimes G$.

Answer 1

Here is my interpretation. Let us consider a category $\mathcal C= \mathcal T\mathrm{op}\times \mathrm{Ab}$. We have two functors, $F,F':\mathcal C\to \mathrm{Ab}$ given by $F(X,G)=C_n(x;G)$ and $F'(X,G)=C_n(X)\otimes G$ (in fact, by taking all $n$ at once, we can view these as taking values in chain complexes). The assertion is that there is a natural transformation between these functors, and that this transformation is is an isomorphism.