 Solution

One form of the Mumford rigidity lemma says that if $X$ is a complete variety, $S$ is connected, and a morphism $F:X\times S\to Z$ maps $X\times \{s_0\}$ to one point, then $F$ factors through the projection to $S$. In particular, if $F$ also maps $\{x_0\}\times S$ to that point, then $F$ is constant.

Choose $s_0\in S(k)$ and put $g=f{|}_{X\times \{s_0\}}$. To see that the pointed morphism $g:X\to Y$ is a homomorphism, apply rigidity toIt vanishes on $X\times \{e\}$, so it factors through the second projection; it also vanishes on $\{e\}\times X$, so it is identically zero. Now defineThen $F(x,s_0)=e$ for every $x$ and $F(e,s)=e$ for every $s$. Rigidity forces $F$ to be identically $e$, so