Hello I have the following problem:
We call a group action $\psi:G\times Y\rightarrow Y$ even if forall $y\in Y$ we have that there is an open neighbourhoow $U_y$ such that $$\psi_g(U_y)\cap U_y\neq \emptyset \Leftrightarrow g=id\,\,\,\,\,\,\,\,\,(1)$$ If $G$ acts evenly on a topological space $Y$ via $$\psi:G\times Y\rightarrow Y$$ then also the group $H\subset G$ acts evenly via the restriction of $\psi$ to $H$
I wanted to proof this as follows:
Let us assume that $G$ acts evenly on $Y$ via $$\psi:G\times Y\rightarrow Y$$ i.e. forall $y\in Y$ there exists $U_y$ such that (1) holds. Let $H\subset G$ a subgroup. Let $y\in Y$ and take $U_y$ as above and consider $$\psi_g:Y\rightarrow Y,\,\,\,y\mapsto g\cdot y$$
$\Leftarrow$ Let us assume that $g=id$. Since $H\subset G$ is a subgroup also $id\in H$. Then from (1) $$\psi_{id}|_H(U_y)\cap U_y=id\cdot U_y\cap U_y\neq \emptyset$$
$\Rightarrow$ Now let us assume that $h\neq id$, then from (1) $$\psi_{h}|_H(U_y)\cap U_y=h\cdot U_y\cap U_y=\emptyset$$ Therefore $H$ acts evenly on the restriction
But I'm not sure if this works. Could someone please help me, I would be very thankful since I have a lot of problems with this topic.
Note that the open neighbourhoods $U_y$ in your definition need to be non-empty. Otherwise $$\operatorname{id}U_y \cap U_y = U_y\cap U_y = U_y = \varnothing.$$
Anyway, it holds that ${\psi}{\mid_H}_{h}(U_y) = \psi_h(U_y)$ for all $h\in H$. Thus $${\psi}{\mid_H}_{\operatorname{id}}(U_y) = \psi_{\operatorname{id}}(U_y)=\operatorname{id}U_y = U_y$$
Therefore $${\psi}{\mid_H}_{\operatorname{id}}(U_y)\cap U_y = \psi_{\operatorname{id}}(U_y)\cap U_y \not= \varnothing$$ by $(1)$.
Now assume $\operatorname{id}\not= h \in H$. As before,
$${\psi}{\mid_H}_{h}(U_y) = \psi_{h}(U_y)=hU_y$$
Thus
$$\psi_{h}(U_y)=hU_y \cap U_y = \varnothing$$ again, due to $(1)$. Right, otherwise there would exist some $h\in H\subset G$ so that $(1)$ is not satisfied. Which would contradict your assumption of $G$ acting evenly on $Y$.