# Why does this homomorphism correspond to a Galois covering?

I need some help with understanding of this formula.

Frobenius formula $N(G;C_1,\ldots,C_k)$ counts the number of $$\{(c_1,\ldots,c_k) \in C_1 \times C_2\times\cdots\times C_k\mid c_1\cdots c_k=1\}$$ for arbitrary conjugacy class $C_1,\ldots,C_k$ in terms of the characters of the irreducible representations of $G$.

Now there is a topological intepretation of this formula, which I've found in the book Graphs on Surfaces and Their Applications.

Let $X$ be a $2$-sphere with $k$ points say $P_1,P_2,\ldots,P_k$ removed, then $\pi_1(X)$ is a free group on $k$ generators $x_1,x_2,\ldots,x_k$ with $x_1x_2\cdots x_k=1$ , and $N(G;C_1,\ldots,C_k)$ simply counts the number of homomorphisms $f$ from $\pi_1(X)$ to $G$ with $f(x_i)\in C_i$ for each $i$.

Then it says, if $G$ acts faithfully on some finite set $F$, then each such homomorphism correspond to a Galois covering of $X$ with fibre $F$, Galois group $G$, and ramification points $P_i$ such that for each $i$, the permutation of the elements of a fixed fibre induced by the local monodromy at $P_i$ belongs to the conjugacy class of $C_i$. Hence $N(G;C_1,\ldots,C_k)$ counts the coverings with these properties.

I cant figure out why each such homomorphism correspond to a Galois covering with these properties ... Can anyone explains it to me ? Thanks.

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Try please to use more spaces and more lines jumps, so as the reading will be smoother. –  DonAntonio Jan 2 at 13:19