Computation of the fundamental group of the projective plane without Van Kampen theorem. I want to find a way of proving $\pi_1(\mathbb R \mathbb P^2)=\mathbb Z_2$ without invoking the Van Kampen theorem. 
My thoughts: I want to think of the real projective plane as $D^2$ with antipodal points on the boundary glued. It suffices to show that there are only two types of loops up to homotopy since $\mathbb Z_2$ is the only group with two elements. But I find it hard to show that there is only one nontrivial loop up to homotopy.
Any approach without using Van Kampen theorem is appreciated!
 A: Prove that $\mathbb{RP}^2 = \frac{\mathbb{S}^2}{\mathbb{Z}_2}$ thinking of $1$ acting as the identity and $-1$ as the antipodal map. Next, prove that the projection $p:\mathbb{S}^2 \rightarrow \mathbb{RP}^2$ is a covering map. Now, you don't have to go through too much of covering spaces to know that whenever you have a covering map $p: E \rightarrow X$, the group $\pi_1(X, x_0)$ acts on the fiber $p^{-1}(x_0)$ as follows:
For $e \in p^{-1}(x_0)$ and $[\sigma] \in \pi_1 (X, x_0)$, $e \cdot [\sigma]:= \sigma_e(1)$, where $\sigma_e$ is the only lifting of the loop $\sigma$ to $E$ starting at $e$.
Now more generally, for an action of a discrete group $G$ on a space $Y$ and $y\in Y$, if we define $G_y$ as the subset of $G$ fixing $y$ (it is a subgroup), then you have a bijection $b:\frac{G}{G_y}\rightarrow G\cdot y$ such that $b \circ q = L_y$ (or $=R_y$), where $q:G \rightarrow \frac{G}{G_y}$ is the natural projection, and $L_y (R_y): G \rightarrow G\cdot y$ is left (right) multiplication by $y$ (the action could be defined as right or left). If $G$ was a topological group then $b$ would be continuous and onto, giving $\frac{G}{G_y}$ to quotient topology.
What happens is that the subgroup of $\pi_1 (X, x_0)$ fixing $e$ is exactly $p_*(\pi_1(E,e))$, where $p_*$ is the map induced by $p$ in the homotopy groups (also not hard to prove, uses uniqueness of liftings with fixed initial condition). Since $\mathbb{S}^2$ has $\pi_1=0$ and this is a two-fold covering, you have a bijection from $\pi_1(\mathbb{RP}^2, \pi(x_0))$ to a set of two elements. The only group with two elements is $\mathbb{Z}_2$.
I recommend you Hatcher's book, or Greenberg's, to review some covering spaces. This technique can be generalized for any space that is quotient by the action of a group, particularly it provides one of the shortest and more comprehensible proofs that $\pi_1(\mathbb{S}^1,*)=\mathbb{Z}$.
