I'm wanting to say that if $V$ is open and $g$ is a group action acting on this set is it true that $g(V)$ is open?

Under theorem 8.1.2 of the following http://www.math.toronto.edu/mat1300/covering-spaces.pdf

this seems to be suggested. It makes sense, but intuition can be deceptive often in mathematics.

  • $\begingroup$ If the group is not acting continuously, then there is no reason for this. In any case, is $V$ the whole space on which the group acts? $\endgroup$ Nov 12 '16 at 18:17
  • $\begingroup$ If the group action is continuous, yes, because then each action is also a homeomorphism. $\endgroup$ Nov 12 '16 at 18:17
  • $\begingroup$ Given that the link you've given is about covering spaces, maybe you have something more specific in mind about what a group action is? $\endgroup$ Nov 12 '16 at 18:19

The group actions as defined in that article are topological group actions, hence continuous.

The map:

$$\phi_g:x\mapsto gx$$

is a homeomorphism - it is a continuous map that has a continuous inverse:

$$\phi_{g^{-1}}:x\mapsto g^{-1}x$$

This means, in particular, that if $V$ is an open subset of $X$ then $\phi_g(V)=gV$ is an open subset of $X$.

  • $\begingroup$ Thank you for your answer. Is my attempt below correct too? I want to phrase it in my own language so that I really know I understand it, thank you for your time. $\endgroup$
    – Clclstdnt
    Nov 12 '16 at 19:55

Thank you Thomas Andrews.... put another way, we have the following:

$\phi_{g^{-1}} = (\phi_g)^{-1}$ since $$(\phi_{g^{-1}}\cdot\phi_g)x=\phi_{g^{-1}}(gx) = g^{-1}gx=x$$ And $$(\phi_{g}\cdot\phi_{g^{-1}})x=\phi_{g}(g^{-1}x) = gg^{-1}x=x$$

And since each action is continuous it follows that $\phi_{g^{-1}}$ is continuous and hence $\phi_{g}$ is open.


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