I'm reading Steenrod The topology of fiber bundle and at the page 7 is this statment that I don't understand.
Namely the last proposition. Take $\phi: G\to\text{Aut}(Y)$ given by $\phi(g)(y)=gy.$ If you take $\phi(g)=\phi(h)$ with the fact that the action is effective we get that $g=h$, so $\phi$ is injective. How do I prove the fact that $\phi$ is surjective?
I think this is a grammatical thing that maybe isn’t perfectly clear. When the author writes “$G$ is isomorphic to a group of homeomorphisms of $Y$” they mean the following:
Here is another way to look at it:
You read the definitions and a topological group is essentially a group where the operation is continuous. And a topological transformation group is basically a group action where everything is continuous. Recall the definition of a group action of $G$ on $X$ is a homomorphism $\phi : G \to \text{Sym}\,X.$
Now we have something like a group action except that the group elements transform the set in a continuous way, thus we get a (made up name) continuous representation: $\eta' : G \to \text{Aut}\,Y,$ where this is the group of topological (ie homeomorphisms) automorphisms.
$G$ being effective corresponds to $\eta'$ having a trivial kernel and then the authors statement follows from the first group isomorphism theorem.
However one should note that our “continuous representation” has forgotten the property in the original definition of $\eta$ being continuous in $G$ as well as $Y$. If $Y$ is locally compact and Hausdorff then $\eta$ being continuous is the same as giving $\text{Aut}\,Y$ the compact-open topology and having $\eta'$ be continuous.