TITLE: Why does the group law commute with morphisms of elliptic curves?
QUESTION [18 upvotes]: I know this should be pretty simple, but right now the only way I can see how to prove it is to sit down and write out explicit formulae for the group law, and see that everything works out. What's the geometric or abstract-nonsense reason why the abelian group structure of elliptic curves behaves nicely under homomorphisms?

REPLY [12 votes]: There is also a simple analytic proof which is enlightening in a different way.
Let $f : E \to E'$ be a morphism of elliptic curves. Both E and E' have $\mathbb C$ as their universal covering space. The composition $\mathbb C \to E \to E'$ lifts uniquely to a continuous map $\overline f$ from $\mathbb C$ into the the universal covering of $E'$ (since the source is simply connected), and $\overline f$ is holomorphic because the map $\mathbb C \to E'$ is locally biholomorphic. 
Let $E = \mathbb{C}/\Lambda$, $E' =  \mathbb{C}/\Lambda'$. We know that for $\lambda \in \Lambda$ we have $\overline f(z+\lambda) - \overline f(z) \in \Lambda'$, hence $\overline f(z+\lambda) - \overline f(z)$ is constant as a function of z. Differentiating, the derivative of $\overline f$ is a doubly periodic entire function, so it is a constant. Since $f$ should preserve the basepoint on the elliptic curves, which is the image of 0, we can assume that $\overline f (0) = 0$ and so $\overline f (z) = cz$ for some constant $c$. But the group laws on E and E' are induced by the group law on $\mathbb C$, so the fact that f preserves the group structure reduces to the distributive law!
The same proof holds also for any complex torus.