Setup. Let $k$ be an algebraically closed field of characteristic zero, and let $G/k$ be a semi-abelian variety i.e., $G$ is a commutative algebraic group which is an extension of an abelian variety $A/k$ by a torus $T/k$, so it admits the following presentation: $$0 \to T \to G \to A \to 0$$ Let $Z/k$ be a smooth, integral variety and let $U\subset Z$ be a dense open such that $\text{codim}(Z\setminus U) \geq 2$.

Question. Is it true that any morphism of $k$-schemes $U\to G$ uniquely extends to a morphism $Z\to G$?

This result is stated in Lemma A.2 of Mochizuki's Topics in absolute anabelian geometry I: generalities and the proof goes as follows. First, he deduces the result when $G$ is proper i.e., $G$ is isomorphic to $A$. In this case the result is well-known and can be proved in a few different ways; one avenue of proof uses that $A$ does not contain any rational curves. Next, he says that we may reduce to the case where $G$ is isomorphic to a torus. Again, this result is well-known and follows from an application of Serre's normality criterion.

I understand the proofs of both cases, but I am struggling to see how to combine these to get the result for a general semi-abelian variety $G/k$. More precisely, I do not understand how one may reduce to the case where $G$ is a torus.

Any comments, suggestions, references, and/or counter-examples would greatly be appreciated!

Setup. Let $k$ be an algebraically closed field of characteristic zero, and let $G/k$ be a semi-abelian variety i.e., $G$ is a commutative algebraic group which is an extension of an abelian variety $A/k$ by a torus $T/k$, so it admits the following presentation: $$0 \to T \to G \to A \to 0$$ Let $Z/k$ be a smooth, integral variety and let $U\subset Z$ be a dense open such that $\text{codim}(Z\setminus U) \geq 2$.

Question. Is it true that any morphism of $k$-schemes $U\to G$ uniquely extends to a morphism $Z\to G$?

This result is stated in Lemma A.2 of Mochizuki's Topics in absolute anabelian geometry I: generalities and the proof goes as follows. First, he deduces the result when $G$ is proper i.e., $G$ is isomorphic to $A$. In this case the result is well-known and can be proved in a few different ways; one avenue of proof uses that $A$ does not contain any rational curves. Next, he says that we may reduce to the case where $G$ is isomorphic to a torus. Again, this result is well-known and follows from an application of Serre's normality criterion.

I understand the proofs of both cases, but I am struggling to see how to combine these to get the result for a general semi-abelian variety $G/k$. More precisely, I do not understand how one may reduce to the case where $G$ is a torus.

Edit. The answer to the question is yes, the result is true and actually holds in a more general setting by a theorem of Weil (see Bosch–Lütkebohmert–Raynaud's Néron Models, Theorem 4.4.1.). I would still like to understand how Mochizuki proposes to deduce the result for a general semi-abelian variety from only knowing this in the cases of an abelian variety and a torus.

Thanks!