Let $k$ be an algebraically closed field, let $S$ be a smooth projective surface over $k$, let $C$ be a smooth projective curve over $k$.
Let $f:S\to C$ be a genus $1$ fibration, i.e., $f$ is a surjective morphism with connected fibers, such that for all but finitely many $v\in C$, $f^{-1}(v)$ is a smooth curve with genus $1$. Let $\sigma:C\to S$ be a section for $f$, i.e., $f\circ \sigma=\text{id}_C$.
I would like to prove the following assertion:
Every fibre intersects $\sigma(C)$ transversally in a single point.
The single point part is easy. If $v\in C$, then $\sigma\left (C\right )\cap f^{-1}(v)=\left \{ \sigma(v) \right \}$. I cannot prove that the intersection is transversal. Moreover, it should be "obvious", since I know this to be true because of these following sources:
First, if you read "Mordell-Weil Lattices" by Matthias Schütt and Tetsuji Shioda, you find it in the proof of Proposition 5.4: The global sections of $f:S\to C$ are in a natural one-to-one correspondence with the $K$-rational points of $E$. Here $K=k(C)$ and $E$ is the generic fiber of $f$.
The proof begins this way:
There we find the assertion (I highlighted the important part).
Moreover, if we read the article "On the Mordell-Weil Lattices" by Tetsuji Shioda, we find it as an "obvious" lemma.
How would you solve it?
Best Answer
Let $p \in F \cap \sigma(C)$ be an intersection point of a fibre and a section. Let $c \in C$ be the unique point such that $\sigma(c)=p$.
The equation $f \circ \sigma = \operatorname{id}_C$ then gives
$$ df_p \circ d \sigma_c = \operatorname{id}_{T_c C}$$
which shows that $df_p$ is injective on the subspace $d \sigma_c(T_c C) = T_p (\sigma(C))$. On the other hand, $\operatorname{ker} df_p$ contains $T_pF$ since $f$ is constant on $F$. So this shows that $$ T_pF \cap T_p (\sigma(C)) = \{0\};$$ in other words the intersection is transversal.