Degree of a proper holomorphic map of Riemann surfaces

($f$ is a closed map.) Let $A\subseteq X$ be closed and let $y\mathrm{\notin }f(A)$. Choose a compact neighbourhood $K$ of $y$, possible by local compactness [F4], and put $K^{\circ }$ for its interior. Since $f$ is proper, $f^{-1}(K)$ is compact, so $A\cap f^{-1}(K)$, a closed subset of it, is compact, and its image $f(A\cap f^{-1}(K))$ is compact, hence closed in $Y$, and does not contain $y$. Then $V:=K^{\circ }\setminus f(A\cap f^{-1}(K))$ is open, contains $y$, and is disjoint from $f(A)$: for $a\in A$ with $f(a)\in V\subseteq K$ one has $a\in f^{-1}(K)$, so $f(a)\in f(A\cap f^{-1}(K))$, which $V$ avoids. Hence $Y\setminus f(A)$ is open and $f(A)$ is closed.

($f$ is an open map.) Let $O\subseteq X$ be open and $x\in O$. Choose a chart at $x$ and a chart at $f(x)$, and let $F$ be the chart expression on a connected neighbourhood $W\subseteq O$ of $x$; by [F3] $F$ is not constant, so the planar open mapping theorem [F3] applied to $F$ on $W$ shows that $F(W)$ is open in the target chart plane, that is, it contains a neighbourhood of the image of $x$. Since charts carry neighbourhoods to neighbourhoods, $f(O)$ contains a neighbourhood of $f(x)$; as $x\in O$ was arbitrary, $f(O)$ is open.

($f$ is onto.) The image $f(X)$ is nonempty, open by step 1.2, and closed by step 1.1; $Y$ is connected, so its only nonempty clopen subset is $Y$ itself and $f(X)=Y$.

(Fibres are finite.) Let $y\in Y$. By properness $f^{-1}(y)$ is compact, and it is nonempty by step 2.1. Each $x\in f^{-1}(y)$ is isolated in $f^{-1}(y)$: in the coordinates of [F1] the fibre over $y$ corresponds, near $x$, to the solutions of $z^e=0$ in a disc, and $0$ is the only solution. Hence $f^{-1}(y)$ is discrete and compact, hence finite; write $f^{-1}(y)=\{x_1,\dots ,x_r\}$ and $e_j:=e_{x_j}(f)$.

(Local constancy of the weighted count.) Fix $y$ and use the notation of step 3.1. For each $j$ choose centred charts as in [F1] and apply [F2] to the chart expression $F_j$, which has $F_j(0)=0$ and ${\mathrm{deg}}_0{F}_j=e_j$: shrink to open neighbourhoods $U_j$ of $x_j$, pairwise disjoint, and $V_j$ of $y$ with $f(U_j)=V_j$, such that for every $y^{\mathrm{\prime }}\in V_j\setminus \{y\}$ the fibre of $y^{\mathrm{\prime }}$ meets $U_j$ in exactly $e_j$ points, and the fibre of $y$ meets $U_j$ only in $x_j$. The set $X\setminus (U_1\cup \cdots \cup U_r)$ is closed, and its image under the closed map $f$ of step 1.1 is closed and does not contain $y$, so there is an open neighbourhood $V\subseteq V_1\cap \cdots \cap V_r$ of $y$ with $f^{-1}(V)\subseteq U_1\cup \cdots \cup U_r$. For $y^{\mathrm{\prime }}\in V\setminus \{y\}$ the fibre lies in ${\bigcup }_j{U}_j$ and meets $U_j$ in exactly $e_j$ points; in the coordinates of [F1] a point $x\in U_j$ with $f(x)=y^{\mathrm{\prime }}\mathrm{\ne }y$ satisfies $z\mathrm{\ne }0$, where $z$ is the source coordinate, and there the derivative of the expression $z^{e_j}$ is $e_j{z}^{e_j-1}\mathrm{\ne }0$, so $e_x(f)=1$ by [F1]; hence $d(y^{\mathrm{\prime }})={\sum }_j{e}_j=d(y)$.

(The degree.) By step 4.1 every point of $Y$ has an open neighbourhood on which $y\mapsto d(y)$ is constant. Fix $y_0\in Y$ and put $D:=\{y\in Y:d(y)=d(y_0)\}$. Then $D$ is nonempty, it is open because step 4.1 gives a neighbourhood of each of its points on which $d$ is constant, and $Y\setminus D$ is open because for $y\mathrm{\notin }D$ step 4.1 gives a neighbourhood of $y$ on which $d$ is constant, with value $d(y)\mathrm{\ne }d(y_0)$, hence disjoint from $D$. So $D$ is a nonempty clopen subset of the connected surface $Y$, hence $D=Y$ by [F6]; that is, $d$ is constant. Its value $d$ is a positive integer because every fibre is nonempty and the sum over the finite fibre of the positive integers $e_x(f)$ is positive and finite.

(Branch values are locally finite.) In the situation of step 4.1, let $x^{\mathrm{\prime }}$ be a critical point of $f$ with $f(x^{\mathrm{\prime }})\in V$. Then $x^{\mathrm{\prime }}\in U_j$ for some $j$, and in the coordinates of [F1] the chart expression is $z\mapsto z^{e_j}$, whose derivative vanishes in a disc around $0$ only at $z=0$ when $e_j\ge 2$ and nowhere when $e_j=1$; a point with $z\mathrm{\ne }0$ therefore has $e=1$ by [F1]. So the critical points above $V$ are among $x_1,\dots ,x_r$, and the branch values in $V$ are among $f(x_1),\dots ,f(x_r)$: finitely many. Hence every point of $Y$ has a neighbourhood containing only finitely many branch values.

(Covering off the branch locus.) Let $y\in Y$ not be a branch value, so $e_x(f)=1$ for every $x\in f^{-1}(y)$ by definition of the branch locus, and let $V$, $U_j$ be as in step 4.1. For each $j$ the chart expression is $z\mapsto z$ on $U_j$ in the coordinates of [F1], so $f{\mathrm{\mid }}_{U_j}$ is a biholomorphism onto $V_j$ with the coordinate change as inverse; restricting to $f^{-1}(V)\cap U_j$ shows that $V$ is evenly covered with $d$ sheets $f^{-1}(V)\cap U_j$, one for each of the $d$ points of the fibre (each contributing $e_j=1$). Hence $f$ is a finite-sheeted covering map of degree $d$ over the complement of the branch locus.

(Conclusion.) Steps 2.1, 3.1, 5.1, 6.1 and 5.2 establish all four claims; when $Y$ is compact, finitely many relatively compact open sets cover $Y$ and each contains only finitely many branch values, so the branch locus is finite. Every selection above was made from the finite fibre of a fixed point and from finitely many charts, so no choice principle is used.