Property of Complete Variety

I have a question about a step in the proof from Lang's "Abelian Varieties" (page 20):

By definition an abelian variety $$A$$ over field $$k$$ is a proper smooth $$k$$-group scheme that is irreducible.

In the Theorem 1 where we have to show that an abelian variety is commutative the author says in the red tagged line that it suffice to show that

$$\dim T \le \dim A$$

where $$T$$ is the locus if $$(x, yxy^{-1}) \in A \times A$$.

Following step isn't clear:

Why completeness of $$A$$ implies that the point $$(e,a) \in T \cap (e \times A)$$ has a preimage $$(e,b)$$ in $$A \times A$$ under the map $$(x,y) \mapsto (x, yxy^{-1})$$?

I'm working with wiki's definition of completeness.

• I think T is a priori Zariski closure of the locus you describe, not the locus itself. Since A is complete, AxA is, i.e. projection onto it from a product of varieties is a closed map. Now, AxA is also separated, meaning that the graph of a morphism from it to itself is a closed subset of AxAxAxA. This means that the locus you describe is closed (being the image of a closed set under a closed map), so a posteriori T is the locus you describe. I think this solves your question (though maybe I am mistaken, from this perspective the proof in the book is somewhat contrived). – Aknazar Kazhymurat Mar 8 at 21:35