Isomorphism between free abelian group and infinite cyclic group generated by identity

Question

I am currently working through John M. Lee's textbook Introduction to Topological Manifolds but have come across a question that has confused me a little. The exercise is below:

Exercise 9.16. Prove that for any set $S$, the identity map of $S$ induces an isomorphism between the free abelian group on $S$ and the direct sum of infinite cyclic groups generated by elements of $S: \mathbb{Z} S \cong \bigoplus_{\sigma \in S} \mathbb{Z}\{\sigma\}$.

I can see that the required isomorphism would have the form:

$$\Psi:\mathbb{Z} S \rightarrow \bigoplus_{\sigma \in S} \mathbb{Z}\{\sigma\}, \quad \Psi \bigg(\sum_{i=1}^k n_i\sigma_i\bigg)= \{f_\sigma\}_{\sigma \in S} \text{ where }f_\sigma = \begin{cases} n_i\sigma_i & \text{ if }\sigma=\sigma_i \text{ appears in the sum}\\ 0 & \text{ otherwise}\\\end{cases}.$$

With inverse given by:

$$\Psi^{-1}:\bigoplus_{\sigma \in S}\mathbb{Z}\{\sigma\} \rightarrow \mathbb{Z} S , \quad \Psi (\{f_\sigma\}_{\sigma \in S})= \sum_{\sigma\in S}f_{\sigma}$$

And my thinking was that the way you would prove the result is to:

1. Define $\Psi$ and $\Psi^{-1}$ as above.
2. Apply either
   • The characteristic property of the coproduct from Theorem 5.57 in the textbook (noting that the direct sum is the coproduct in the category of abelian groups), to show that $\Psi^{-1}$ is a homomorphism from $\bigoplus_{\sigma \in S} \mathbb{Z}\{\sigma\}$ to $\mathbb{Z} S$
   • Or the characteristic property of free abelian groups (Proposition 9.14 (a)) to show that $\Psi$ is a homomorphism from $\mathbb{Z} S$ to $\bigoplus_{\sigma \in S} \mathbb{Z}\{\sigma\}$.
3. Conclude $\Psi$ is a bijective homomorphism and, thus, an isomorphism, and the result follows.

My confusion is this: the question says we must consider the isomorphism induced by the identity map on $S$. I don't understand what the identity map on $S$ has to do with this problem. Please could someone explain where the answer should be referencing the identity map of $S$ and why? I feel I've misunderstood something obvious...

Answer 1

We are dealing with two groups $\mathbb{Z}S$ and $\bigoplus_{\sigma\in S} \mathbb{Z}\sigma$, generated by $S,$ in the sense that every element of the group has the unique representation with respect to the set $S.$ Formally a fixed element $s\in S$ in the second group is identified with the element of the form $\bigoplus_{\sigma\in S}\sigma(s)$ where $\sigma(s)=s $ if $\sigma=s$ and $\sigma(s)=0$ for $\sigma\neq S.$ But informally it is natural to treat this element as $s.$ In this sense the mapping $s\mapsto \bigoplus_{\sigma\in S}\sigma(s)$ is informally called the identity and can be extended to the group isomorphism.