Consider a smooth projective curve $X$ over $\mathbb C$ (so $X$ is a projective $\mathbb C$-scheme, integral, of finite type....), and let $t: X\longrightarrow\mathbb P^1_{\mathbb C}=\operatorname{Proj}(\mathbb C[T_0,T_1])$ be a finite morphism of degree $d$. By the known equivalence between smooth projective curves and compact Riemann surfaces, there is a holomorphic map $t(\mathbb C): X(\mathbb C)\longrightarrow\mathbb P^1(\mathbb C)$ of compact Riemann surfaces, associated to $t$.
Now my question(s):
Is $t(\mathbb C)$ a $d$-branched covering of the projective line? In general do finite morphisms of degree $d$ in scheme theory setting correspond to $d$-branched coverings in the classical framework of compact Riemann surfaces?
Reasoning by analogy I think that the answer is yes to both questions but I'm not formally sure of this.
Thanks in advance.
Just to have the question answered.
Under the correspondence between irreducible projective smooth curves over $\mathbb{C}$, and Riemann surfaces, function fields are preserved. More precisely, $K(X)\cong M(X^\text{an})$, and that this isomorphism is functorial.
Now, if $f:X\to Y$ is finite of degree $d$, then $[K(X):K(Y)]=d$. Thus, for the associated map $f^\text{an}:X^\text{an}\to Y^\text{an}$ we have that $[M(X^\text{an}):M(Y^\text{an})]=d$.
It is then a common fact of complex analysis that if $f:X\to Y$ is map of compact Riemann surfaces, then the degree as a branched covering is $[M(X):M(Y)]$. While there is probably a more canonical reference, the one that comes to mind is Szamuely's Galois Groups and Fundamental Groups, Proposition 3.3.5.