Let $E:y^2z=x^3+Axz^2+Bz^3$ be an elliptic curve over a field $K$. Considering the usual abelian group $(E, +)$, I'm trying to prove the following for $m\in\mathbb{N}\setminus\{0\}$:
The multiplication map $[m]:E(\overline{K})\to E(\overline{K})$ given by $P\mapsto mP=P+...+P$ is surjective.
I already know how to prove the following:
($1$) The addition map $+:E(\overline{K})\times E(\overline{K})\to E(\overline{K})$ is a morphism of projective varieties.
($2$) $[m]$ is not constant.
Here is my attempt to prove surjectivity:
[I'll say "projective variety" meaning irreducible projective variety implicitly]
Proof: By (1), $+:E(\overline{K})\times E(\overline{K})\to E(\overline{K})$ is a morphism of projective varieties, so $[2]= id+id$ is a morphism and so is $[m]=id+[m-1]$ by induction. Since morphisms of projective varieties map closed subsets to closed subsets (and are also continuous), $Im([m])$ is a subvariety of $E(\overline{K})$. Since $\dim E(\overline{K})=1$, we get:
$$Im([m])\subsetneq E(\overline{K})\Rightarrow\dim Im([m])=0\Rightarrow Im([m]) \text{ is a point }$$
By (2), we know $Im([m])$ cannot be a point, so $Im([m])=E(\overline{K})$.
Is this proof correct?