Suppose $S_1$ and $S_2$ are two finite-type hyperbolic Riemann surfaces, meaning they are finite genus compact Riemann surfaces with a finite number of punctures, and both have $\mathbb{H}$ as their universal cover. For example, a sphere with 4 punctures or a torus with 1 puncture are such spaces. Let us write $\Sigma_1$ and $\Sigma_2$ for the corresponding compact surfaces, so $S_i = \Sigma_i\setminus A_i$ where $A_i\subset \Sigma_i$ is finite.
Suppose that $S_1$ has finite hyperbolic area, and $f: S_1\to S_2$ is a nonconstant holomorphic map. It's a theorem [1, Corollary 3] that such a map admits a holomorphic a extension $\tilde f:\Sigma_1\to\Sigma_2$. This map is necessarily surjective.
Now, it's my belief that since $S_1$ and $\Sigma_1$ have finite area, then $\Sigma_2$ (and hence $S_1$) have finite area. Is this correct? And if so, what would be the best way to prove it? Do you need to look at $\tilde f$ at all, or is it automatically true that a finite-type hyperbolic surface has finite area? Something about this statement feels off to me.
For context, I am trying to show that a certain finitely-generated torsion-free Fuchsian group $\Gamma\leq {\rm PSL}(2,\mathbb{Z})$ has finite index in the modular group. I believe that, to do this, it suffices to show that the corresponding quotient surface $S_2 = \mathbb{H}/\Gamma$ has finite area.
My knowledge in this area is very sketchy so references are appreciated too.