Cayley complex of $\langle a|a^2\rangle$ is a covering of $\mathbb{R}P^2$ In Hatcher Example 1.47, we constructed the $2$-fold cover of $\mathbb{R}P^2$ by finding the Cayley complex of $G=\langle a|a^2\rangle$. I understand the construction and how the $2$-cells are attached, but I am worried about whether the action is free. It's a fact that the left translation action on $Cay(G,S)$ is free if and only if $S$ doesn't contain any elements of order $2$. In this case, consider the edge $\{e,a\}$ in $Cay(G,\{a\})$ and then $a\cdot \{e,a\}=\{a,e\}$ which is a fixed point.
Would someone like to remind me what's wrong here?
 A: There are several different definitions of graphs. The one you are using (as said in your comment) might be called an abstract simplicial complex of dimension 1. It is not the appropriate definition for working with Cayley graphs nor Cayley 2-complexes, because by your definition there are no loop edges and no bigons, whereas a
Cayley graph generally allows both loops and bigons.
In particular, the graph in this problem, i.e. the Cayley graph of the group presentation $G = \langle a \mid a^2 \rangle$, has a bigon.
So I strongly recommend that you come to terms with the topological definition of a CW complex, in which vertices are actual points in the space, and edges are actual subsets of the space a list of topological properties, and so on for higher dimensional cells. The Cayley 2-complex is always a CW complex. And while you might not actually need the full definition of CW complex right away in your studies, you'll need it very soon.
In the situation of your post, the Cayley 2-complex of $G$ can be denoted by simply enumerating the two 0-cells as $v_0$, $v_1$, and the two 1-cells as $e_0,e_1$, and two 2-cells as $f_0,f_1$. For each $i$ modulo 2 the boundary of the 1-cell $e_i$ is homeomorphic to $[0,1]$ and its boundary is $\{v_0,v_1\}$. For each $i$ modulo 2 the boundary of the 2-cell $f_i$ is homeomorphic to a closed 2-dimensional disc and its boundary is $e_0 \cup e_1$.
Under this description the Cayley 2-complex is homeomorphic to $S^2$.
The action of the nontrivial element $a$ is $a \cdot v_i = v_{i+1}$ (using arithmetic modulo 2) and similarly $a \cdot e_i = e_{i+1}$ and $a \cdot f_i = f_{i+1}$. This is conjugate to the antipodal self-map of $S^2$ and so the quotient is indeed $\mathbb RP^2$.
