Definition (Polynomial-like mapping) Let $U$ and $V$ be open sets of $\mathbb{C}$ conformally equivalent to $\mathbb{D}$ such that $\overline{U}\subset V$ and let $f:U\to V$ be a proper holomorphic map such that every point in $V$ has exactly $d$ preimages in $U$ when counted with multiplicity. The triple $(f,U,V)$ is called a polynomial-like mapping of degree $d$.
In that case, $f:U\to V$ is a branched covering map with $d-1$ critical points in $U$ when counted with multiplicity.
It seems that the critical values of $f$ are the only singular ones because $V$ is bounded (then there are not asymptotic values), so my question refers to the critical values of an holomorphic function.
If I am not wrong, it is as if a critical value of degree $d-1$ leads to exactly $d$ different branchs for the inverse $f^{-1}$ arround it. Althought maybe it is obviously, I can not seen it and I would like a result that justifies it or some kind of explanation.
Example $f(z) = z^n$: The critical value has degree $n-1$, so its "inverse" has $n$ different branches.
Thank you.
First off, I think you are missing in your definition of polynomial like map that $f$ should be proper.
Secondly, your question as stated does not make sense. There is no local inverse near a critical value; that's the point. A branch of $f^{-1}$ at $z$ is the choice of a local inverse map $g$ defined on some neighborhood $U$ of $z$ such that for all $x \in U$, $f(g(x))=x$.
What is true however is that if $c$ is a critical point of order $d-1$ (meaning that $f(z)=f(c)+a (z-c)^d+O\left((z-c)^{d+1}\right)$ for some $a \neq 0$) and $v=f(c)$ is the corresponding critical value, then $f$ has local degree $d$ at $c$ (meaning that there is an open neighborhood $U$ of $c$ such that for all $y \in f(U)$ with $y \neq v$, $y$ has $d$ preimages by $f$ in $U$.
Edit : a bit more details
Up to replacing $f$ by $g(z)=\frac{1}{a}(f(z-c)-v)$, we may assume that $f$ is of the form $f(z)=z^d+O(z^{d+1})$. Then the claim is that on a neighborhood of $0$, there exists a holomorphic function $\phi$ such that $f(z)=\phi(z)^d$. To construct that function $\phi$, note that $f(z)=z^d (1+h(z))$ with $h(z)=o(1)$, so that we may define $\phi(z)=z \exp(\frac{1}{d} \mathrm{Log} (1+h(z)))$. This is well defined whenever say $|h(z)|<1$, which is the case near $z=0$. The claim follows easily from the existence of $\phi$.