The genus of an algebraic curve is invariant under isomorphisms I would like to know how to prove (or even better to see a full proof) of the following "fact".


Let $C_1$ and $C_2$ be two smooth curves and let $\phi : C_1 \rightarrow C_2$ be an isomorphism. Then
    $$
\text{genus}(C_1) = \text{genus}(C_2)
$$


I am not completely sure this is true since I haven't seen this result explicitly stated, but I Imagine it has to be true. 
The motivation for this comes from an exercise from Silverman's book The Arithmetic of Elliptic Curves. I was doing the following exercise and I found that I needed the above mentioned fact in order for my argument for $(i) \implies (ii)$ to work.


2.5 Let $C$ be a smooth curve. Prove that the following are equivalent (over $\bar{K}$):
(i) $C$ is isomorphic to $\mathbb{P}^1$.
(ii) $C$ has genus $0$.
(iii) There exist distinct points $P, Q \in C$ satisfying $(P) \sim (Q)$


I've thought about it but unfortunately I don't really see how to easily relate the dimensions of the Riemann Roch spaces associated to each curve. 
I would really appreciate some help with this. 
Thank you.
 A: The notation $\text{genus}(C_1)$ doesn't even make sense unless you know that the genus is invariant under isomorphism; if it isn't, the genus must depend on information other than $C_1$ which you haven't provided. 
In any case, the definition of genus you are given implies that it is unique, and since the various dimensions $\ell(D)$ are defined independently of any choices they are automatically invariant under isomorphism, so the definition you have been given already comes with a guarantee that $g$ is invariant under isomorphism. But if you want a "proof" anyway, then setting $D = 0$ gives $\ell(K_C) = g$, so it suffices to show that the canonical divisor is invariant under isomorphism (that's why it's called the canonical divisor!).
A: The genus of a curve, viewed as the dimension of the space of holomorphic form $H^0(C,\Omega_C^1)$ is even invariant under birational morphisms. Indeed, if $f:X \to Y$ is a birational morphism between smooth complex projective varieties, then $f^*$ induces an isomorphism $H^0(Y,K_Y^{\otimes m}) \to H^0(X,K_X^{\otimes m})$ for each integer $m \geqslant 0$, where $K_X = \Lambda^{\dim X} T_X^*$ is the canonical bundle on $X$.
