Preamble: I want to understand why the number of irreducible representations of a finite group equals the number of conjugacy classes of the group. The only direct proof I managed to find is in the book Representation Theory, A First Course, where the authors prove in the proposition 2.30 that the number of irreducible representations of a group $G$ is equal to the number of conjugacy classes of the group $G$. The authors then state that this is equivalent for the characters $\{\chi_V\}$ forming an orthonormal basis for $\mathbb{C}_{\text{class}}(G)$ , where $\mathbb{C}_{\text{class}}(G)$ is the set of class functions $f:G\to \mathbb{C}$ such that $\forall h \in G:f(hgh^{-1}) = f(g)$ that is, the class functions are constant on conjugacy classes. The given proof is the following
Suppose that $\alpha:G\to \mathbb{C}$ is a class function and $(\alpha,\chi_V) = 0$ for all irreducible representations $V$; we must show that $\alpha = 0$. Consider the endomorphism $\varphi_{\alpha, V} = \sum\alpha(g) \cdot g:V\to V$. By Schur's lemma $\varphi_{\alpha, V} = \lambda \mathrm{Id}$; and if $n = \mathrm{dim}(V)$, then $\lambda = \frac{1}{n}\mathrm{tr}(\varphi_{\alpha, V}) = \frac{1}{n}\sum\alpha(g)\chi_V(g) = \frac{|G|}{n}\overline{(\alpha, \chi_{V^*})} = 0$. Thus $\varphi_{\alpha, V} = 0$, or $\sum\alpha(g)\cdot g = 0$ on any representation $V$ of $G$; in particular this will be true for the regular representation $V = R$. But in $R$ the elements $\{g \in G\}$, thought of as elements of $\mathrm{End}(R)$, are linearly independent. For example, the elements $\{g(e)\}$ are all independent. Thus $\alpha(g) = 0$ for all $g$, as required.
Problem: The proof above is nice and concise; my only problem with it is that I don't understand how it shows that number of conjugacy classes of $G$ = number of irreducible representations of $G$. Since I've just began learning representation theory, I think that a more direct/elementary proof suits me better
Alternative approach: One old representation theory exam asked essentially the same question by letting $A$ be the set of all conjugacy classes of $G$ and $B$ the set of all irreducible complex representations of $G$. Then, we are given an orthogonality relation $\sum_{a \in A}\frac{|a|}{|G|}\overline{\chi_{f}(a)}\chi_g(a) = \begin{cases}1 & f \cong g\\0 & f \not\cong g\end{cases}$, where $f, g \in B$, and $a$ is a representative of a conjugacy class and $|a|$ is the number of elements in the class. Our task is then to show that for $a_1, a_2 \in A:\sum_{f \in B}\overline{\chi_f(a_1)}\chi_f(a_2) = \begin{cases}\frac{|G|}{|a_1|}&: a_1 = a_2\\0 &: a_1 \neq a_2\end{cases}$
Main question: A given hint in the problem is that we should define a linear map from $\mathbb{C}^A = \{f:A\to \mathbb{C}\}$ to $\mathbb{C}^B = \{f:B\to \mathbb{C}\}$. I defined the function $\alpha_f(\chi_g) = \overline{\chi_f}\chi_g, f, g \in B$ and showed that this map is linear. I also reasoned why the map is injective: $\alpha(g) \equiv 0$ on $G$ means that either $\overline{\chi_f} \equiv 0$ which is impossible as $f$ is an irreducible representation, or $\chi_g \equiv 0$, which I suppose is possible if $\chi_g$ is the zero scalar multiple of some other character. Currently, I'm stuck at showing that $\alpha$ is surjective. Namely, I don't know how to reverse-engineer that any irrep. function has a pre-image in $\mathbb{C}^A$.