Let $G$ be a discrete group and $X$ be a Tychonoff $G$space. Then there exists a $G$action on StoneČech compactification $\beta X$. If the fixed point set $X^{G}\neq \emptyset $, then the StoneČech compactification of the fixed point set is the fixed point set of the StoneČech compactification, i.e., $\beta \left( X^{G}\right) =\left( \beta X\right) ^{G}$?

1$\begingroup$ $\beta$ is a left adjoint functor, and taking $G$fixed points is a limit. So I suspect this isn't true as in general left adjoint don't preserve limits. $\endgroup$– Niall TaggartOct 23, 2018 at 10:12

$\begingroup$ To start with, one has a canonical continuous map $\beta(X^G)\to (\beta X)^G$. Also the assumption $X^G\neq\emptyset$ is irrelevant as it can be artificially enforced by adding an isolated fixed point to $X$. $\endgroup$– YCorOct 23, 2018 at 22:51
1 Answer
Not without further assumptions. First create an ordered space $X$ by identifying $\langle0,\omega_1\rangle$ and $\langle1,\omega_1\rangle$ in the product $2\times(\omega_1+1)$ to a point, $\Omega$ say. Then make the product $X\times X$ and take out $\langle\Omega,\Omega\rangle$. You can rotate the resulting space $Y$ around that hole over $\pi$ by mapping $\bigl<\langle i,\alpha\rangle,\langle j,\beta\rangle\bigr>$ to $\bigl<\langle 1i,\alpha\rangle,\langle 1j,\beta\rangle\bigr>$ and similarly for points with $\Omega$ as a coordinate. This map $f:Y\to Y$ has no fixed points but $\beta Y=X$ and $\beta f$ does have a fixed point: $\langle\Omega,\Omega\rangle$.
Now take the sum $Y\cup\{0\}$ and extend $f$ by $f(0)=0$. Then $f$ has one fixed point and $\beta f$ has two. As $f$ is its own inverse this gives us an action of $\mathbb{Z}/2\mathbb{Z}$ that is a counterexample.