Numerical equivalence of divisors on fibered surface Let $C$ be a smooth projective curve over $\mathbb{C}$ of genus $g > 2$ and $\pi_i\colon C\times C \to C$ the projection onto the the $i$-th factor. Let $f_i \in \mathrm{Num}(C\times C)$ be the classes of the fibers of the $\pi_i$ and $\delta \in \mathrm{Num}(C\times C)$ the class of the diagonal. If we set $\delta^\prime = \delta - f_1 - f_2$, we have $f_1 \cdot f_2 = 1$, $f_i^2 = f_i \cdot \delta^\prime = 0$ and $(\delta^\prime)^2 = -2g$.
Let $[D] = d_1 f_1 + d_2 f_2 + b \delta^\prime$ be the class of an irreducible curve $D\subset C\times C$ such that $\pi_1\restriction_D$ is an isomorphism (i.e., $d_2 = 1$) and $d_1 \geq 1$. Why do we have $[D] = \delta$ if $\mathrm{Aut}(C) = \lbrace\mathrm{id}\rbrace$? And why $[D] \in \lbrace \delta, 2f_1 + 2f_2 - \delta \rbrace$ if $C$ is hyperelliptic and $\mathrm{Aut}(C) = \lbrace \mathrm{id}, \iota \rbrace$ where $\iota$ is the hyperelliptic involution?
 A: First let's show that $\pi_2$ restricted to $D$ is also an isomorphism. If it was not, then since $d_1 \geq 1$ it would give a finite map $D \to D$. But since $g>2$, the Euler characteristic $\chi(D)$ is negative, and then Riemann–Hurwitz shows that no such map exists. So in fact $\pi_2$ must be an isomorphism. 
So now the map $\alpha := \pi_2 \circ \pi_1^{-1}$ is an automorphism of  $C$. But this map sends $x \in C$ to the unique point $y(x)$ such that $(x,y(x)) \in D$. 
So if $\text{Aut}(C)$ is trivial, we must $\alpha=\text{id}$, hence $y(x)=x$; that is, $D$ consists of pairs $(x,x)$ — in other words, $D$ is the diagonal (even as a curve, not just a numerical class).
If $\text{Aut}(C)=\{\text{id},\iota\}$, then either $\alpha=\text{id}$, in which case $D$ must be the diagonal as before, or else $y(x)=\iota(x)$. So in the latter case, $D$ consists of pairs $(x,\iota(x))$. 
To calculate the numerical class of $D$ in the latter case, intersect it with curves representing $f_1$, $f_2$, and $\delta$. We get
$$[D] \cdot f_1 = [D] \cdot f_2 = 1 \\ [D] \cdot \delta =2g+2$$ 
We get the last line by noting that $D$ intersects $\delta$ precisely at fixed points of $\iota$, and these can be counted by Riemann–Hurwitz.
Now write $[D]=af_1+bf_2+c\delta$, and use the numbers above to solve for $a$, $b$, and $c$.
