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3 added 10 characters in body

It seems that Ben is nevertheless right that the answer to the first question is "NO". Let $G=SL(2,\Bbb C)$, and $B$ be the subgroup of lower triangular matrices. Then the inclusion $B\to G$ is an epi, since every algebraic representation of $B$ that extends to $G$ does so uniquely (on the nose, not just up to an isomorphism!). This follows from the fact that in any finite dimensional representation $V$ of the Lie algebra $sl(2)$, the operator $e$ is determined by $f$ and $h$. Indeed, the kernel $K$ of $e$ is spanned by vectors $v$ satisfying $hv=mv$ and $f^{m+1}v=0$ for some integer $m\ge 0$, and since $V=\Bbb C[f]K$, the operator $e$ on $V$ is uniquely determined.

So one might guess that a morphism of complex affine algebraic groups $\phi: H\to G$ is an epi if and only if $G/\phi(H)$ is connected and proper (but I did not check this).

2 added 176 characters in body

It seems that Ben is nevertheless right that the answer to the question is "NO". Let $G=SL(2,\Bbb C)$, and $B$ be the subgroup of lower triangular matrices. Then the inclusion $B\to G$ is an epi, since every algebraic representation of $B$ that extends to $G$ does so uniquely (on the nose, not just up to an isomorphism!). This follows from the fact that in any finite dimensional representation $V$ of the Lie algebra $sl(2)$, the operator $e$ is determined by $f$ and $h$. Indeed, the kernel $K$ of $e$ is spanned by vectors $v$ satisfying $hv=mv$ and $f^{m+1}v=0$ for some integer $m\ge 0$, and since $V=\Bbb C[f]K$, the operator $e$ on $V$ is uniquely determined.

So one might guess that a morphism of complex affine algebraic groups $\phi: H\to G$ is an epi if and only if $G/\phi(H)$ is connected and proper (but I did not check this).

1

It seems that Ben is nevertheless right that the answer to the question is "NO". Let $G=SL(2,\Bbb C)$, and $B$ be the subgroup of lower triangular matrices. Then the inclusion $B\to G$ is an epi, since every algebraic representation of $B$ that extends to $G$ does so uniquely (on the nose, not just up to an isomorphism!). This follows from the fact that in any finite dimensional representation $V$ of the Lie algebra $sl(2)$, the operator $e$ is determined by $f$ and $h$. Indeed, the kernel $K$ of $e$ is spanned by vectors $v$ satisfying $hv=mv$ and $f^{m+1}v=0$ for some integer $m\ge 0$, and since $V=\Bbb C[f]K$, the operator $e$ on $V$ is uniquely determined.