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?



Why does the group law commute with morphisms of elliptic curves?
This follows by a rigidity property of certain morphisms. It is important to note that elliptic curves are complete, that is, proper and integral schemes. Then we have the following "Rigidity Lemma" (see Mumford's Abelian Varieties, the beginning of chapter II (page 43 of the old edition), for example):


Let $X$ be a complete variety, $Y$ and $Z$ any varieties, and $f:X\times Y\rightarrow Z$ a morphism such that for some $y\_0\in Y$, $f(X\times\{y\_0\})$ is a single point $z\_0$ of $Z$. Then there is a morphism $g:Y\rightarrow Z$ such that if $p\_2:X\times Y\rightarrow Y$ is the projection, $f=g\circ p\_2$.


How does this help? Well, elliptic curves have a distinguished point (the origin), and a morphism of elliptic curves is a morphism of their underlying schemes taking one distinguished point to the other. We'd like to show that any such morphism is actually compatible with the group law. So consider a morphism of elliptic curves $f:E\_1\rightarrow E\_2$, and let $\Phi:E\_1\times E\_1\rightarrow E\_2$ be defined by $\Phi(x,y)=f(x+y)-f(x)-f(y)$. Then $\Phi(E\_1\times\{0\})=0$ so by the lemma I quoted, there is a morphism $g:E\_1\rightarrow E\_2$ such that $\Phi=g\circ p\_2$. And since $\Phi(\{0\}\times E\_1)=0$, as well, $g$ must be zero. But then $\Phi$ itself must be zero.


None of this relied on $E\_1$ and $E\_2$ being elliptic curves; it works exactly the same way for abelian varieties.


