In a topological semigroup $G$, multiplying by a fixed element is continuous as we can decompose it into $$G \overset{\text{diagonal}}{\longrightarrow} G\times G \overset{\text{const}\times\text{id}}{\longrightarrow} G\times G \overset{\text{m}}{\longrightarrow} G,$$ in which all the component functions are continuous.
However, I am curious whether this is an open map.
i.e., given any open set $U\subseteq G$ and $g\in G$ , is $gU$ open?
If it is not in general, how about inverse semigroups or regular semigroups?
This is true for groups, but it fails for inverse semigroups. Take $G = \{0, 1\}$ with the usual multiplication of integers. Then $G$ is a topological inverse semigroup for the trivial topology. Indeed, the open sets are $\emptyset$ and $G$ and if $f: G \times G \to G$ is the map defined by $f(x,y) = xy$, then $f$ is continuous since $f^{-1}(G) = G \times G$. However, $G$ is open, but $0G = \{0\}$ is not.