When studying connected components in matrix groups, we consider a closed subgroup $S$ of $GL_n(k)$ ($k$ algebraically closed field), where 'closed' refers to the fact that we can consider $GL_n(k)$ as an affine variety over $k$ endowed with the Zariski topology (in which affine varieties are exactly the closed sets).
Take $g\in S$ fixed. I want to show that left multiplication $\phi:S\to S: y\mapsto gy$ defines a homeomorphism. This map is clearly bijective. As for continuity, we would have to take a closed subset $V$ in $S$ (which is an affine subvariety) and show that $\phi^{-1}(V):=\{s\in S: gs\in V\}$ is closed (i.e., an affine variety) as well.
Let $V=V(I)$ for some radical $I\trianglelefteq k[S]$, then $\phi^{-1}(V) = \{s\in S: f(gs)=0, \forall f\in I\}$. How do I get this in the form $V(J)$ for some ideal $J$?
Here's what I think is a cleaner approach.
First, left multiplication by $g$ is a morphism $S\to S$: matrix multiplication is given by polynomials, so it's a morphism $M_n(k)\times M_n(k)\to M_n(k)$. Now restrict to the closed subvariety $\{g\}\times S$, and use the fact that $S$ is a subgroup to conclude that the restriction lands in $S$.
Next, note that left multiplication by $g$ is invertible: just left-multiply by $g^{-1}$. As this is also a morphism (by the same reasoning in the above paragraph), we see that $g\cdot -$ and $g^{-1}\cdot -$ are mutually inverse morphisms $S\to S$ (aka they are both isomorphisms). As a morphism is continuous with respect to the Zariski topology, $g\cdot -$ is a homeomorphism.