I was working on questions about computing fundamental groups from past qualifying exam papers and I wanted to know if I'm going about it correctly. The question reads
Recall that the standard embedding of $\mathbb{R}P^1$ in $\mathbb{R}P^2$ is the image of the equator under the 2-fold cover $\pi : S^2 \to \mathbb{R}P^2$ given by $\pi (x) = \pi (-x)$. Let $X$ be the union of two projective planes glued via the identity map of the standardly embedded $\mathbb{R}P^1$s. Compute the fundamental group of $X$.
My idea was to consider $\mathbb{R}P^2$ as an upper hemisphere with the boundary circle identified by the antipodal map. This boundary circle (when sent to the quotient) is $\mathbb{R}P^1$. Now instead of gluing after sending to the quotient, I thought of gluing before, so that effectively you take two hemispheres, glue them along the boundaries and apply the antipodal map to the boundary circle. Now, applying Seifert-van Kampen, I have one generator $x$ for one copy and another generator $y$ for the other copy, satisfying $x^2 = y^2 = e$. For the amalgamation, the inclusions of the generator of the intersection into either $\mathbb{R}P^2$ are $x^2$ and $y^2$ respectively, so the new relation is $x^2y^{-2} = e$ which is already true. So we get $\pi _1 (X) = \langle x, y | x^2 = y^2 = e \rangle$ which is just $\mathbb{Z}_2 * \mathbb{Z}_2$
Is this correct? Because this is also the fundamental group of $\mathbb{R}P^2 \vee \mathbb{R}P^2$, so I'm kind of unsure.
Not quite. When gluing, you need to keep in mind that the boundary circle has been quotiented out by the antipodal map. This means that the intersection of the two copies of $\mathbb{R}P^2$ is the boundary circle mod the antipodal identification, not the boundary circle itself. In particular, the generator of the fundamental group of the intersection corresponds to just $x$ and $y$, not $x^2$ and $y^2$, since after quotienting by the antipodal map you only have to go around the circle halfway to get a loop. So the amalgamation ends up identifying $x$ and $y$ in $\pi_1(X)$, so you get just $\mathbb{Z}_2$.