Let $f \colon X \to \mathbb P^n$, $n \geq 2$, be a holomorphic map from a compact Riemann surface $X$ and whose image $f(X)$ is a smooth projective curve. There are two notions of degree for such a map $f$:
the degree $d_1$ as the degree of the map of Riemann surfaces $\widetilde f: X \to f(X)$ which is the number of sheets of the ramified covering;
the degree $d_2$ of any divisor in the associated linear system $|f|$, or the degree of the hyperplane divisor $f^*(H)$.
We know that these notions of degree are different and are related: $d_2 = d_1 \cdot \deg f(X)$.
Passing to homology, we see that $f$ induces the following maps:
- $\widetilde f_* \colon H_2(X) \to H_2(f(X))$. Since both groups are isomorphic to $\mathbb Z$, this is given by $\widetilde f_*[1] = k[1]$ and one can show that $k=d_1$ (up to a sign).
- $f_* \colon H_2(X) \to H_2(\mathbb P^n)$. Both groups are again isomorphic to $\mathbb Z$, so this map is $f_*[1] = m[1]$ for some $m$. Is it true that $m = d_2$ (up to a sign)?
Due to the formula relating $d_1$ to $d_2$, we can reformulate the above question in the following form. We have $f = i \circ \widetilde f$, where $i\colon f(X) \hookrightarrow \mathbb P^n$. This map descends to $f_* = i_* \circ \widetilde f_*$, where $i_* \colon H_2(f(X)) \to H_2(\mathbb P^n)$. Since the latter spaces are isomorphic to $\mathbb Z$, we have $i_*[1] = l[1]$. Thus, having $m=d_2$ is equivalent to having $l = \deg f(X)$. Thus, the equivalent question is whether $\deg f(X) = l$ (up to a sign)?