How to prove that a group action is continuous iff actions of each group element are homeomorphisms

Question

Let $X$ be a topological space and $G$ a group, equipped with the discrete topology, with an action on $X$, and define $\phi:G\times X\rightarrow X$ by $\phi(g,x) = g\cdot x$. I'm trying to show that $\phi$ is continuous if and only if each map $\ell_g:X\rightarrow X:x\mapsto g\cdot x$ is a homeomorphism of $X$.

The forward direction is fairly straightforward. The reverse direction (i.e. showing each $\ell_g$ is a homeomorphism) is proving more difficult, and I am starting to wonder if it is even true. We know that each $\ell_g$ is a bijection, and continuity of $\ell_g^{-1}$ follows from continuity of $\ell_g$ for all $g\in G$ by $\ell_g^{-1}$ = $\ell_{g^{-1}}$. The trouble is that it seems very difficult to extract information about a particular $\ell_g$ from the continuity of $\phi$. In particular, if $U$ is open in $X$, we can write

$$\phi^{-1}(U) = \bigcup_{g\in G}\{g\}\times\ell_g^{-1}(U) = \bigcup_{g\in G}\{g\}\times\ell_{g^{-1}}(U) $$

but I see no clear way to move from this to the fact that $\ell_{g^{-1}}(U)$ is open in $U$, which would complete the proof. Any help getting unstuck (or a counterexample) would be appreciated!

Answer 1

I would be very surprised if each $\ell_g: X \rightarrow X$ being a homeomorphism implies that $\phi: G \times X \rightarrow X$ is continuous. It seems almost certainly false to me, so I would be curious to see your proof of this.

Anyway, to prove that the continuity of $G \times X \rightarrow X$ implies that each $\ell_g: X \rightarrow X$ is a homeomorphism. Having fixed $g$, it is clear that the map $\ell_g$ is continuous, since it is a composition of continuous maps

$$X \rightarrow \{g\} \times X \rightarrow G \times X \xrightarrow{\phi} X$$

$$x \mapsto (g,x) \mapsto (g,x) \mapsto g.x.$$

Indeed, the first map is a homeomorphism, the second map is an inclusion of a subspace into a larger space, and the third map you assumed was continuous.

For the same reason, $\ell_{g^{-1}}$ is continuous and it along with $\ell_g$ is a homeomorphism since $\ell_g \circ \ell_{g^{-1}}$ is the identity function on $X$.

Answer 2

In order to prove that $\phi$ is continuous, what has to be proved is that $\phi^{-1}(U)$ is open in $G \times X$ for each open subset $U \subset X$.

For each $g \in G$, the subset $\{g\}$ is open in the discrete topology on $G$.

Also, knowing that $\ell_{g^{-1}} : X \to X$ is continuous, it follows that the subset $\ell_{g^{-1}}(U)$ is open in $X$.

The set $\{g\} \times \ell_{g^{-1}}(U)$ is therefore open in the product topology on $G \times X$.

And since the union of open sets is open, it follows that $\bigcup_{g \in G} \{g\} \times \ell_{g^{-1}}(U)$ is open in $G \times X$. This proves that $\phi$ is continuous.