The <Mumford rigidity lemma> says that if $A$ is complete, $B$ is connected, and a morphism $F:A\times B\to Z$ maps the fiber over some $b_0\in B$ to a point, then $F$ is constant on every $A$-fiber and factors through $B$.

Put $y=f(e_X)$ and normalize
$$
g=T_{-y}\circ f,
$$
so $g(e_X)=e_G$. Define
$$
F:X\times X\longrightarrow G,
\qquad F(x,z)=g(x+z)g(z)^{-1}.
$$
When the first coordinate is $e_X$, this is constantly $e_G$. Apply rigidity with the second copy of the complete variety $X$ as the complete factor. It follows that $F(x,z)$ is independent of $z$, and evaluation at $z=e_X$ gives $F(x,z)=g(x)$. Therefore
$$
g(x+z)=g(x)g(z),
$$
so $g$ is a <homomorphism of group varieties> and $f=T_y\circ g$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 126 / 2 / ii / Solution

A homomorphism of group varieties is a morphism of algebraic varieties that also preserves multiplication. It automatically preserves the identity and inversion.


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 Incoming links: Homomorphism of group varieties