Is the intersection of the diagonal with a graph always transverse in characteristic zero Let X be a projective smooth connected curve over $\mathbf{C}$. Let $f:X\to X$ be a non-constant morphism.
Is the intersection of the diagonal $\Delta_X$ and the graph $\Gamma_f$ on $X\times X$ transversal? That is, do we have that $\Delta_X \cdot \Gamma_f = \# \mathrm{Fix}(f)$, where $\mathrm{Fix}(f)$ is the set of fixed points of $f$?
This is not true in positive characteristic. Consider the morphism $(x:y)\mapsto (y:x)$ from $\mathbf{P}^1$ to itself. In characteristic two, there is precisely one fixed point $(1:1)$ with multiplicity $2$, i.e., $\Gamma_f\cdot \Delta_X = 2$. (In characteristic $\neq 2$, there are precisely two fixed points with multiplicity 1: $(1:1)$ and $(1:-1)$.)
I also tried to do it with an elliptic curve, but the example I wrote down also gave a transversal intersection.

The map $(x:y)\mapsto(xy+x^2:y^2)$ on $P^1$ works, no?
