A surjective morphism of abelian varieties induces an epimorphism on the torsion subgroups

Question

Let $f:A\to B$ be a surjective morphism of abelian varieties (over an algebraically closed field).

Why is it true that $f$ induces an epimorphism on the points of finite order $A_{\mathrm{tors}}\to B_{\mathrm{tors}}$?

Answer 1

Clearly the image of a torsion point by a group homomorphism is again a torsion point, so it only remains to prove the surjectivity of the restriction $f \colon A_{\mathrm{tors}} \to B_{\mathrm{tors}}$.

By using Stein factorization, we see that $f$ factors as a surjective homomorphism having an abelian variety as kernel and an isogeny, so it sufficies to prove the result in these two situations.

$\boldsymbol{(i)}$ Assume first that the kernel of the surjective homomorphism $f \colon A \to B$ is an abelian subvariety. Let $b \in B_{\mathrm{tors}}$ be a point of order $n$, namely $nb=0$. Then if $a \in A$ is any preimage of $b$ via $f$, we have $$f(na)=nf(a) =nb=0,$$ hence $na \in \ker f$.

By assumption $\ker f$ is an abelian subvariety of $A$, in particular it is a divisible group; so we can find $a' \in \ker f$ such that $na = na'$, that is $n(a-a')=0.$ So $a-a' \in A_{\mathrm{tors}}$ and moreover $$f(a-a')=f(a)-f(a') = b-0= b,$$ i.e. $a-a'$ is a preimage of $b$ that lies in $A_{\mathrm{tors}}$. This shows the desired surjectivity in this case.

${}$ $\boldsymbol{(ii)}$ Assume now that $f \colon A \to B$ is an isogeny; then $\ker f$ is finite, in particular it is a torsion subgroup of $A$. Choosing $b$ and $a$ as above, we see that $$f^{-1}(b)=a+ \ker f.$$ Now $na \in \ker f$ and the fact that $\ker f$ is a torsion subgroup imply that $a$ is a torsion element, too. So in this case all the preimages of $b$ via $f$ are torsion elements of $A$, in particular the surjectivity of $f \colon A_{\mathrm{tors}} \to B_{\mathrm{tors}}$ follows also in this case.