Topological groups acting on sets

Question

Let $G$ be a topological group and $X$ be a topological space. Consider the following continuous action

\begin{align*} \alpha: G\times X &\to X \\ (g,x) &\mapsto gx. \end{align*}

I am trying to justify why following map is also continuous \begin{align*} \alpha_x: G &\to X \\ g &\mapsto gx. \end{align*}

If $A$ is open on $X$ then the inverse image $\alpha^{-1}(A)$ is open in $G\times X$. Then it is a union of elements of the form $U\times V$, where $U$ is open on $G$ and $V$ is open on $X$. In particular, the union of those $U$'s is open in $G$ and it is the inverse image $\alpha_x^{-1}(A)$.

Can someone tell me if this argument is correct?

Answer 1

Your argument is correct, modulo the weird/wrong $\alpha^{-1}=(A)$ notation.

It really resolves to the fact that the map $i_x:G\to G\times X$ defined as $g\mapsto (g,x)$ is continuous, and $\alpha_x=\alpha\circ i_x$ is the composition of continuous functions.

Why is $i_x$ continuous? You've essentially proven this in your proof, but it also results from the "universal property" of the product topology:

Any map $f:A\to B\times C$ is continuous if and only if $\pi_1\circ f: A\to B$ and $\pi_2\circ f:A\to C$ is continuous.

In the above case, $A=B=G, C=X,$ $\pi_1\circ i_x$ is the identity function on $G$ so is continuous. And $\pi_2\circ i_x$ is the constant function $g\mapsto x,$ so is continuous.

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In general, given any continuous $f:B\times C\to D,$ and any $x\in C,$ we have that the function $f_x:B\to D$ defined as $b\mapsto f(b,x)$ is continuous for the same reasons. There is nothing necessary about $\alpha$ being a group action, only that $\alpha$ is continuous.