How can I show the action of left translation by a topological group $G$ on itself is a continuous action using the ordinary definition of continuity?

Question

Let $G$ be a topological group and let the action $f: G\times G \to G$ be defined by $f(g,g')=gg'$.

I want to show that this action by left translation is a continuous action by showing that if $U \subset G$ is open, then $f^{-1}(U)$ is open in $G\times G$.

I have seen that it is continuous since it is the composition $G\xrightarrow{i_g} G\times G\xrightarrow{m}G$, where $i_g(g')=(g,g')$ and where $m$ is group multiplication.

How can we show that it is continuous using the ordinary definition of continuity?

Answer 1

"Proving it" would be a bit weird, as $f=m$ and so $f$ is continuous by very definition of a topological group.

Answer 2

We wanted proved that $i_g(g):G→G\times G$ is continuous, that is, if A is a open in topology on $G\times G$ then $i_{g}^{-1}(A)$ is open in topology on G.

In other ways, note that $i_g(g'):G→f(g)\times h(g)$, where $f$ is a constant function ($f(x)=g$, for all x $\in G$) and $h(x)=x,$ for all $x \in G$.

Given $f$ and $h$ are continuous function, then $i_g(g')$ is continuous function