I am trying to calculate the fundamental group of $\mathbb{RP}^n$ without using the Van Kampen Theorem. My basic idea consists of the following: First of all, $\mathbb{RP}^1\simeq S^1$ and thus, the fundamental group for $n=1$ is just $\mathbb {Z}$.
For $n\geq 2$ I have thought of the following covering map: $ p:S^n\to S^n/_{x\sim-x}\simeq \mathbb{RP}^n,\ x\mapsto [x]$. This is obviously continuous and surjective. Furthermore, $S^n\simeq \mathbb{RP}^n\times \{x,-x\}$. Thus, this map should be a covering.
Now we look at the function $\phi: \pi_1(\mathbb{RP^n},x_0)\to p^{-1}(x_0)=\{x_0,-x_0\},\ [w]\mapsto L(w,x_0)(1)$ where $w$ is a loop in $\mathbb{RP}^n$ and $L$ is the lift of $w$. Now, it is known that $\phi$ is well-defined and bijective. Now we know that $p^{-1}(x_0)\simeq \mathbb{Z}_2$. Thus, in order to prove that the fundamental group of $\mathbb{RP}^n$ is indeed $\mathbb{Z}_2$, we need to show that $\phi$ is a group homomorphism. This is where I am struggling at. Could someone help me?
We will proof that $\pi_1(\mathbb{R}\mathbb{P}^n) \tilde{=} \mathbb{Z}$ for $n = 1$ and $\pi_1(\mathbb{R}\mathbb{P}^n) \tilde{=} \mathbb{Z} / 2 \mathbb{Z}$ for $n \geq 2$:
To establish the assertion for $n=1$, we show that $S^1 / \sim$ is homeomorphic to $S^1$. Thereby, we think of $S^1$ as a subset of $\mathbb{C}$. Let $\psi : S^1 \to S^1$ be the map $\psi(z) := z^2$. Clearly, $\psi$ is continuous and surjective. We have \begin{equation} z \sim w \iff \psi(z) = \psi(w). \end{equation} Thus there is a bijection $\tilde{\psi} : S^1/\sim \to S^1$, such that $\psi = \tilde{\psi} \circ p_1$. Since $S^1 / \sim$ is endowed with the final topology, $\tilde{\psi}$ is continuous. Since $S^1 / \sim$ is compact and $S^1$ is Hausdorff, $\tilde{\psi}$ is a homeomorphism.
To treat the case $n \geq 2$ we are going to analyse the fundamental group $\pi_1(S^n/ \sim)$ using a covering given by the space $S^n$ and covering projection $p_n$. We get this covering the following way:
For $n \geq 2$ we define $\sim \subseteq S^n \times S^n$ be the equivalence relation \begin{equation} x \sim y :\iff (x=y \lor x = -y) \end{equation}
and let $p_n : S^n \to S^n / \sim$ be the canonical projection. Let $S^n / \sim$ be endowed with the final topology $\mathcal{V}$ induced by $\lbrace p_n \rbrace$. Then $(S^n , \mathcal{T}_n, p_n)$ is a covering of $(S^n/\sim, \mathcal{V})$.
To start with, let us fix notation. Set $x_0 := (1,0, \ldots, 0) / \sim \in S^n / \sim$, which will be used as base point and set $\tilde{x_0} := (1,0, \ldots, 0)$. For a loop $f \in L(S^n / \sim, x_0)$ denote by $\tilde{f}$ the lifting of $f$ with initial point $\tilde{x_0}$.
Now we use that the covering $p_n : S^n \to S^n / \sim$ is 2-fold. For every loop $f \in L(S^n / \sim, x_0)$ the terminal point of a lifting $\tilde{f}$ belongs to the two-element set $\lbrace \tilde{x_0}, - \tilde{x_0}\rbrace$. Now we use that $S^n$ is simply connected. If $f,g \in L(S^n / \sim, x_0)$ with $\tilde{f}(1)= \tilde{g}(1)$, then $\tilde{f} \sim \tilde{g}$ and hence $f \sim g$. We conclude that $\pi_1(S^n/ \sim)$ has at most two elements.
Let $1_{x_0}$ be the constant path at $x_0$. Its lifting $\tilde{1_{x_0}}$ clearly is the constant path at $\tilde{x_0}$. If $f$ is a loop with $f \sim 1_{x_0}$, then by the following theorem:
we have $\tilde{f}(1) =\tilde{x_0}$. In order to show that $\pi_1(S^n/\sim) \neq \lbrace 1 \rbrace$, it is thus sufficient to find one loop $f$ with $\tilde{f}(1) \neq \tilde{x_0}$.
Consider the path $\tilde{f} : [0,1] \to S^n$ given as $\tilde{f}(t) := (\cos(\pi t), \sin(\pi t), 0, \ldots, 0)$ for $t \in [0,1]$. Then $\tilde{f}(0) = \tilde{x_0}$ and $\tilde{f}(1) = - \tilde{x_0}$. Pushing $\tilde{f}$ to $S^n / \sim$ gives a loop with base point $x_0$: $f(t) := (p_n \circ \tilde{f})(t) = (\cos(\pi t), \sin(\pi t), 0, \ldots, 0)/ \sim$ for $t \in [0,1]$.
By definition $\tilde{f}$ is the lifting of $f$.
To conclude the proof, note that up to ismorphism there is only one group with two elements, namely $\mathbb{Z} / 2 \mathbb{Z}$.