The Covering Transformation Theorem - Fund. Group of the Proj. Plane

# The Covering Transformation Theorem - Fundamental Group of the Projective Plane

Recall from The Covering Transformation Theorem for Finding Fundamental Groups page that if $X$ is a topological space and $(\tilde{X}, p)$ is a universal cover of $X$ then:

(1)
\begin{align} \quad A(\tilde{X}) \cong \pi_1(X, x) \end{align}

(Where $A(\tilde{X})$ denotes the set of covering transformations of $\tilde{X}$.)

We will now look at applying this result to find the fundamental group of the projective plane, $\mathbb{P}^2$.

Let $X = \mathbb{P}^2$. Let $\tilde{X} = S^2$ and let $p : S^2 \to \mathbb{P}^2$ be defined for each $x \in S^2$ by:

(2)
\begin{align} \quad p(x) = [x] \end{align}

Where $[x]$ denotes the equivalence class of $x$ obtained from the quotient space of the sphere $S^2$ by identifying antipodal points on $S^2$. Then $(S^2, p)$ is the universal cover of $\mathbb{P}^2$.

Let $f : S^2 \to S^2$ be defined for all $x \in S^2$ by $f(x)$ to be the antipodal point of $x$ on $S^2$. Then the set of all covering transformations of $S^2$ (with respect to $p$) is:

(3)
\begin{align} \quad A(\tilde{X}) = \{ \mathrm{id}_{\tilde{X}}, f \} = \langle f \rangle \cong \mathbb{Z}_2 \cong \pi_1(X, x) = \pi_1(\mathbb{P}^2, x) \end{align}

So the fundamental group of $\mathbb{P}^2$ is isomorphic to $\mathbb{Z}_2$.