Why there is no meromorphic function of degree $d=1$ on any compact Riemann surface of positive genus? We have the Riemann-Hurwitz formula:
$$
2g_X-2=d(2g_Y-2)+\sum_{x\in X}(e_x-1)
$$
It is said that from this we can deduce that there is no meromorphic function of degree $d=1$ on any compact Riemann surface of positive genus.
I wonder how?
If I let $d=1$, I can get
$$
2(g_X-g_Y)=\sum_{x\in X}(e_x-1)
$$
but what's next? Maybe I lack some knowledge about meromorphic function on Riemann surface, anyone can help?
 A: Let $f:X\to \mathbb C$ a meromorphic function or equivalently a holomorphic function $f:X\to \mathbb P^1$ . Suppose d=1. Then f is bijective holomophic map and therefore a biholomorphic map. Follow that $g(X)=g(\mathbb P^1)$ because $g$ is a topological invariant.
another way:
Like you said : $$2( g(X)-g(\mathbb P^1))=\sum(e_x-1)$$
but $e_x=1$ for all x because $d=1$ and as we know $g(\mathbb P^1)=0$
Therefore $g(X)=0$
A: The degree of a map between compact Riemann surfaces is known to be a constant.
Recall that, locally around $P\in X$ and $f(P)=Q$, such maps look like $z\mapsto z^e$, where $e=e_P$ is unique.
The degree $d$ being constant is equivalent to the following:
$$\sum_{P\mapsto Q} e_P=d.$$
Thus, since $e_P\geq 1$, one cannot have ramification at any point if $d=1$.  On the other hand, this means that $f$ is bijective since $e_P=1$ for every point, and this shows that $f$ is biholomorphic (it is so locally since $e_P=1$, and it is globally since $f$ is bijective).
If you want to see a funny example where $d=1$ you may consider the compactification of the plane curve 
$$y^2=x^2(x+1),$$
which is not a Riemann surface since it has a node at the origin.  The map $t\mapsto (t^2-1, t(t^2-1))$ is generically of degree $1$ (where $t=y/x$), but this fails to work at the origin.
Topologically, the target space looks like a sphere with two points identified, which is therefore not a topological manifold.
If $X,Y$ are as in the question, the degree-$1$ case is rather boring (biholomorphisms and that's that).
Thus, in the case where $g(X)$ and $g(Y)$ differ, no such map exists.
