Prove that for every set there is a free module on that set

Question

While studying the book "Module theory: an approach to linear algebra" by T. S. Blyth (the electronic edition of the book is free to download), I came across a proof that confused me. We need to prove that for every non-empty set $S$ and every unitary ring $R$ there is a free $R$-module on $S$. Here's the complete proof:

First part of the proof second part of the proof

I couldn't grasp the next to last line, specifically, the following equality:

$$h'(v) = \sum_{s \in S}v(s)h'[f(s)]$$

Edit:

It turned out that the source of my confusion was this equality:

$$v = \sum_{s \in S}v(s)f(s)$$

I thought that $f(s)$ is an element of $R$, when in fact $f(s)$ is a mapping (from $S$ to $R$), and therefore the sum will also be a mapping.

Answer 1

It seems that it is just

$$h^\prime(v) = h^\prime\left(\sum_{s \in S}v(s)[f(s)]\right) = \sum_{s \in S}v(s)h'[f(s)]$$

as $h^\prime$ is supposed to be an $R$-morphism.

Answer 2

The equality follows from $R$-linearity of the morphism $h'$, and from the previous equation $$\theta = \sum_{s\in S} \theta(s)f(s)$$ Applying $h'$ to such a sum results in the desired equation.