[proofplan]
The argument is a direct application of the [Riemann–Hurwitz Formula](/theorems/2192). The formula expresses $2g(C) - 2$ as the sum of $d(2g(C') - 2)$ and a non-negative ramification correction $\sum_p (e_p - 1)$, where $d \geq 1$ is the degree of $\phi$. Since the ramification term is non-negative and $d \geq 1$, the formula yields $2g(C) - 2 \geq 2g(C') - 2$ once we verify that the right-hand side is monotone in $d$ when $2g(C') - 2 \geq 0$ and otherwise handle the case $g(C') \leq 1$ separately. The two cases $g(C') = 0, 1$ and $g(C') \geq 2$ are dealt with explicitly so that no sign ambiguity in $2g(C') - 2$ obscures the conclusion.
[/proofplan]
[step:Apply the Riemann–Hurwitz formula to $\phi$]
The morphism $\phi: C \to C'$ is nonconstant between smooth projective curves over $k$, hence the hypotheses of the [Riemann–Hurwitz Formula](/theorems/2192) are met. Let $d := \deg \phi \geq 1$ and let $e_p := e_p(\phi) \in \mathbb{Z}_{\geq 1}$ denote the ramification index of $\phi$ at the closed point $p \in C$. Then
\begin{align*}
2g(C) - 2 = d\bigl(2g(C') - 2\bigr) + \sum_{p \in C} (e_p - 1).
\end{align*}
The sum on the right has only finitely many nonzero terms, since $\phi$ is unramified away from a finite set.
[/step]
[step:Bound the ramification correction term from below by zero]
Each ramification index satisfies $e_p \geq 1$, so $e_p - 1 \geq 0$ for every $p$. Summing the non-negative integers,
\begin{align*}
R := \sum_{p \in C} (e_p - 1) \geq 0.
\end{align*}
Substituting into the Riemann–Hurwitz identity:
\begin{align*}
2g(C) - 2 \geq d\bigl(2g(C') - 2\bigr).
\end{align*}
[/step]
[step:Conclude $g(C) \geq g(C')$ by case analysis on $g(C')$]
We must compare $d(2g(C') - 2)$ with $2g(C') - 2$. We split into three cases according to the sign of $2g(C') - 2$.
**Case 1: $g(C') \geq 2$.** Then $2g(C') - 2 \geq 2 > 0$, and $d \geq 1$ implies
\begin{align*}
d(2g(C') - 2) \geq 2g(C') - 2.
\end{align*}
Combining with the inequality from Step 2:
\begin{align*}
2g(C) - 2 \geq d(2g(C') - 2) \geq 2g(C') - 2,
\end{align*}
so $g(C) \geq g(C')$.
**Case 2: $g(C') = 1$.** Then $2g(C') - 2 = 0$, and the Riemann–Hurwitz inequality from Step 2 reads
\begin{align*}
2g(C) - 2 \geq d \cdot 0 = 0,
\end{align*}
hence $g(C) \geq 1 = g(C')$.
**Case 3: $g(C') = 0$.** Then $g(C') = 0$ and the conclusion $g(C) \geq g(C') = 0$ holds because $g(C) \in \mathbb{Z}_{\geq 0}$ for any smooth projective curve.
In every case $g(C) \geq g(C')$, which is the desired inequality.
[guided]
The case analysis is essential because the prefactor $d$ in $d(2g(C') - 2)$ flips the direction of the inequality depending on the sign of $2g(C') - 2$.
**Why we cannot just write $d(2g(C') - 2) \geq 2g(C') - 2$.** If $2g(C') - 2 < 0$ — that is, $g(C') = 0$ so the value is $-2$ — then multiplying by $d \geq 1$ gives a *more* negative number: $d \cdot (-2) = -2d \leq -2$, so we would get $2g(C) - 2 \geq -2d$ which is *weaker* than the desired $2g(C) - 2 \geq -2$. The naive monotonicity step fails when $g(C') = 0$. This is why Case 3 must be handled by a separate observation — namely, that $g(C) \geq 0$ by definition because the genus is a non-negative integer.
**Case 1 ($g(C') \geq 2$).** Here $2g(C') - 2 > 0$ is a positive integer, and $d \geq 1$ gives $d(2g(C') - 2) \geq 1 \cdot (2g(C') - 2) = 2g(C') - 2$. Together with $R \geq 0$ in Step 2, we get $2g(C) - 2 \geq 2g(C') - 2$, hence $g(C) \geq g(C')$. The strict inequality $g(C) > g(C')$ would require $d > 1$ or $R > 0$ — for example, a hyperelliptic curve of genus $g \geq 2$ admits a degree-$2$ map to $\mathbb{P}^1_k = $ a genus-$0$ curve, for which the inequality is far from tight (it is strict by a huge amount).
**Case 2 ($g(C') = 1$).** Here $2g(C') - 2 = 0$, so the Riemann–Hurwitz formula reduces to $2g(C) - 2 = R \geq 0$, giving $g(C) \geq 1$. There is no contribution from the degree $d$ in this case because the prefactor multiplies zero.
**Case 3 ($g(C') = 0$).** Here $C' \cong \mathbb{P}^1_k$ and $2g(C') - 2 = -2$, so the formula gives $2g(C) - 2 = -2d + R$, i.e. $g(C) = 1 - d + R/2$. Since $R \geq 0$ and $d \geq 1$, we have $g(C) \geq 1 - d + 0$, which can be negative for $d \geq 2$. But $g(C) \geq 0$ is automatic because $g$ is a non-negative integer; this directly gives $g(C) \geq 0 = g(C')$.
**The strategic content.** This theorem says that morphisms can only go from "more complicated" to "less complicated" curves in the sense of genus — a higher-genus curve cannot map nonconstantly to a lower-genus one. The proof is a *single-line* application of Riemann–Hurwitz once the case structure is laid bare. The genuine input is the [Riemann–Hurwitz Formula](/theorems/2192) itself, which extracts the genus arithmetic from the ramification data of $\phi$.
**A common confusion.** Some texts state this theorem as $2g(C) - 2 \geq d(2g(C') - 2)$ and stop, pretending this directly gives $g(C) \geq g(C')$. As Case 3 shows, this is false in general: when $g(C') = 0$, the right-hand side $d(2g(C') - 2) = -2d$ becomes more negative as $d$ grows, weakening the inequality. The true statement $g(C) \geq g(C')$ requires combining the Riemann–Hurwitz inequality with the elementary fact $g(C) \geq 0$.
[/guided]
[/step]
