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Let $f: A \rightarrow B$ be a morphism of abelian varieties defined over a finite field $k$. Let $G$ be a finite group of $A$ and $\pi:A\rightarrow A/G$ the quotient morphism.

Looking at just the group structure, it is enough to have that $G\subset \ker f$ to ensuresensure the existence of a group morphism $g$ such that $f=g\circ \pi$.

When is this $g$ is in fact a morphism of varieties?

Let $f: A \rightarrow B$ be a morphism of abelian varieties defined over a finite field $k$. Let $G$ be a finite group of $A$ and $\pi:A\rightarrow A/G$ the quotient morphism.

Looking just the group structure, it is enough to have that $G\subset \ker f$ to ensures the existence of a group morphism $g$ such that $f=g\circ \pi$.

When this $g$ is in fact a morphism of varieties?

Let $f: A \rightarrow B$ be a morphism of abelian varieties defined over a finite field $k$. Let $G$ be a finite group of $A$ and $\pi:A\rightarrow A/G$ the quotient morphism.

Looking at just the group structure, it is enough to have that $G\subset \ker f$ to ensure the existence of a group morphism $g$ such that $f=g\circ \pi$.

When is this $g$ in fact a morphism of varieties?

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morphism of abelian variety

Let $f: A \rightarrow B$ be a morphism of abelian varieties defined over a finite field $k$. Let $G$ be a finite group of $A$ and $\pi:A\rightarrow A/G$ the quotient morphism.

Looking just the group structure, it is enough to have that $G\subset \ker f$ to ensures the existence of a group morphism $g$ such that $f=g\circ \pi$.

When this $g$ is in fact a morphism of varieties?