I am considering the following situation:
Let $X$ be the space consisting of two circles $A$ and $B$ joined at a point $P$. Let $Y$ be a space that is two circles and four half-circles, joined as shown, and let $p:Y \to X$ be the mapping taking each piece of $Y$ to the correspondingly labeled piece of $X$ as indicated:
The group of deck transformations is defined as
$$\operatorname{Aut}\big(Y/_{X}\big):=\{\varphi:Y\to Y\mid\varphi \text{ a homeomorphism s.t. }p\circ \varphi = p\} $$
In this example, we simply have $\operatorname{Aut}\big(Y/_{X}\big)=\{\operatorname{Id}\}$.
In the lecture, it was mentioned that the group action of $\operatorname{Aut}\big(Y/_{X}\big)$ on $Y$ is not even. I thought this means that $$\forall y \in Y: \exists N \in \mathfrak{N}(y): \varphi_1(N) \cap \varphi_2(N) = \emptyset ~~~ \forall \varphi_1\neq \varphi_2 \in \operatorname{Aut}\big(Y/_{X}\big): \varphi_1\neq \varphi_2 $$ but I must be mistaken.
Does anyone know what it means for $\operatorname{Aut}\big(Y/_{X}\big)$ to act evenly on $Y$?
Thank you in advance!