About representation ring $R(G)$ I have questions about the representation ring $R(G)$ of a finite group $G$. I read the book "Representation Theory. A first Course" written by Fulton and Harris, and there is the following sentence on p.22 which I ask for:

To begin, the character defines a map $$\chi:R(G) \to \mathbb{C}_{\rm class}(G)$$ from R(G) to the ring of complex valued functions on $G$;

How is this map defined ? I think that it is defined by $\chi([V]) = \chi_{V}$ ($V$ is a representation of $G$ and $\chi_{V}$ is $V$'s character), but is this correct?
I have another question. Why is the map $\chi$ injective? I don't have the confidence of the proof for this fact. For now, I prove it as follows:
Let $[V]$ and $[W]$ $\in$ $R(G)$, and let $\chi([V]) = \chi([W])$. By $[V]$ and $[W] \in R(G)$,
$$[V] = \sum a_{i}[V_{i}],$$
$$[W] = \sum b_{i}[V_{i}],$$
where $a_{i}$,$b_{i} \in {\mathbb Z}$, $V_{i}$ is the irreducible representation of $G$. Thus,
$$\chi([V]) = \sum a_{i}\chi_{V_{i}},$$
$$\chi([W]) = \sum b_{i}\chi_{V_{i}}.$$
Since $\chi([V]) = \chi([W])$ and $\{ \chi_{V_{i}} \}$ are linear independent, we obtain $a_{i} = b_{i}$. Therefore, we have $[V] = [W]$. $\Box$
 A: The map is defined as you state. The map $\chi$ maps a representation $(\pi,V)$ to its character $\chi_{V}$.
In your question, the ring $\mathbb{C}_{\mathrm{class}}(G)$ is not the ring of complexed valued functions on $G$. It is actually the ring of so-called "class functions", which means a function on the set of conjugacy classes of $G$.Notice that $\chi_{V}(g)$ only depends on the conjugacy class containing $g$, hence $\chi_{V}$ is of course a class function on $G$.
This map is also a ring homomorphism.
In the representation ring, the addition is defined as the direct sum and the multiplication is defined as the tensor product. So $\chi$ is a homomorphism because we have the following relations about characters:
$$\chi_{U\oplus V}=\chi_{U}\oplus\chi_{V}, \chi_{U\otimes V}=\chi_{U}\cdot \chi_{V}.$$
Then we can prove the injectivity of $\chi$ easily: if $[(\pi,V)]$ is an element in $\ker \chi$, then $\chi_{V}(g)=0$ for each $g\in G$. In particular, $0=\chi_{V}(1)=\mathrm{Tr}(\pi(1))=\dim V$, which implies that $[(\pi,V)]$ must be the zero representation, thus $\chi$ is injective.
