Algebraic group implies quasitopological group

Statement
Any algebraic group naturally acquires the structure of a quasitopological group where the topology is taken to be the Zariski topology. In other words, the following are true:


 * 1) The multiplication map is separately continuous in the two inputs, i.e., it is continuous in each input holding the other input constant.
 * 2) The inverse map is continuous.

Related facts

 * Topological group implies quasitopological group
 * Algebraic group not implies topological group

Proof for the multiplication map
Given: An algebraic group $$G$$.

To prove: The multiplication map $$G \times G \to G$$ is continuous in each variable. More explicitly, for any $$g \in G$$, the multiplication maps $$x \mapsto gx$$ and $$x \mapsto xg$$ are continuous maps from $$G$$ to itself with the Zariski topology.

Proof: We will fix $$g \in G$$ for the proof.

Alternative formulation of the proof for multiplication
Another way of thinking of the proof is as follows. Consider the following three possible topologies on $$G \times G$$:


 * The product topology on $$G \times G$$ arising from the Zariski topology on $$G$$.
 * The Zariski topology arising from the product variety structure on $$G \times G$$ that in turn arises from the variety structure on $$G$$.
 * The coarsest topology on $$G \times G$$ such that any separately continuous map $$G \times G \to G$$ (for the Zariski topology on $$G$$) is continuous from that topology on $$G \times G$$ to the Zariski topology on $$G$$. Explicitly, this is a topology where a subset of $$G \times G$$ is open if its intersection with each fiber of the form $$\{ g \} \times G$$ is open in that copy of $$G$$ and its intersection with each fiber of the form $$G \times \{ g \}$$ is open in that copy of $$G$$.

Proof for the inverse map
The inverse map is an algebraic map by the definition of algebraic group, hence is continuous in the Zariski topology by the definition of Zariski topology.