Question about heights on elliptic curves- the set $\{P \in E(K) : h_f(P) \leq C \}$ is finite (Silverman's AEC)

Question

I am stuck on the proof of Proposition 6.1 here (in Silverman's AEC).

I think the only point I don't get is why $f$ is a finite-to-one map? If I assume this, the rest of the proof follows readily.

My attempt: showing $f$ is finite-to-one is equivalent to showing that if we pick a $Q$ in the set $\mathcal{B} =\{ Q \in \mathbb{P}^1(K) : H(Q) \leq C \}$ then since $f$ is a surjective morphism (yes, any $f\ in K(E)$ induces a surjective morphism $: E \longrightarrow \mathbb{P}^1$), it suffices to show that there are only finitely many choices for $P \in \{ P \in E(K) : h_f(P) \leq C\}$ such that $f(P) = Q$.

I guess we will now suppose that $ P = [X,Y,Z]$ and show finitely many choices for each of $X, Y,$ and $Z$.

$$ H(f(P)) = \prod_{ v \in M_K} max \{ |a|_v , |b|_v \} $$

where $Q = [a,b]$. But I don't know what I should do to get $X,Y,Z$ into the picture as they're what matter here.

It would be great if someone could help me out on this. I'd really appreciate it.

Thank you.

Notation(from Silverman's AEC): $h_f(P) = h(f(P)) = \log H(f(P))$, and in general for any $ P' = [x_0,x_1,... , x_N]$ where $x_i \in K$, $H_K(P') = \prod_{v \in M_K} max\{|x_0|_v, ... , |x_N|_v \}^{n_v}$, $H(P') = H_K(P')^{1/[K:Q]}$.

Answer 1

Are you okay with the fact that we are identifying $f$ with a map $E\to\Bbb P^1$? [If not, go back and look at example $2.2$].

If yes, then this is just the fact that a nonconstant map between curves over $K$ is always finite-to-one. This is a combination of the following two facts [see Theorems $2.4$ and $2.6$]:

If $\phi:C_1\to C_2$ is a nonconstant map of curves defined over $K$, then $\deg(\phi):=[K(C_1):\phi^*K(C_2)]$ is finite.

If $\phi:C_1\to C_2$ is a nonconstant map of smooth curves, then for every $Q\in C_2$ we have $$\sum_{P\in\phi^{-1}(Q)}e_\phi(P)=\deg(\phi),$$ where $e_\phi(P)\ge1$ is the ramification index of $\phi$ at $P$.