How to find fundamental groups and covering spaces of $\mathbb{RP}^2\vee \mathbb{RP}^2$? The following is an exercise I was assigned in homotopy theory.
Defined $X = \mathbb{RP}^2\vee \mathbb{RP}^2$.
a) Find $\pi_1(X)$.
b) Find the universal cover of $X$.
c) Find all of its connected $2$-sheeted covers.
I am trying to use van Kampen's theorem for (a).
I decompose $X$ into two open sets $U_1$ and $U_2$, where each $U_i$ contains one copy of $\mathbb{RP}^2$ as well as a neighborhood around the point of attachment small enough so that each $U_i$ deformation retracts onto the copy of $\mathbb{RP}^2$ that it contains.
Then van Kampen's theorem gives that $\pi_1(X)$ is a quotient of $\mathbb{Z}_2*\mathbb{Z}_2$, where we quotient out by the subgroup normally generated by the elements $\iota_{1, 2}(\omega)\iota_{2, 1}(\omega)^{-1}$, where $\iota_{1, 2}: \pi_1(U_1 \cap U_2) \to \pi_1(U_1)$ is the inclusion map, and similarly for $\iota_{2, 1}$, and the $\omega$'s are arbitrary elements of $\pi_1(U_1 \cap U_2)$.
I want to say that $U_1 \cap U_2$ is simply connected, so each $\iota_{1, 2}(\omega)\iota_{2, 1}(\omega)^{-1}$ is in fact trivial, so then $\pi_1(X)$ is isomorphic to $\mathbb{Z}_2*\mathbb{Z}_2$. Is this true, or did my reasoning go awry somewhere?
For (b), I know that the universal cover is just an infinite chain of copies of $S^2$, where the covering map is just ``locally'' the quotient map.
For (c), I know that the only covering spaces of $\mathbb{RP}^2$ is itself and $S^2$, but I don't really know how to go from there to finding all the connected $2$-sheeted covers.
To summarize: is in fact $\pi_1(X)$ isomorphic to the free product of two copies of $\mathbb{Z}_2$, and how do I find all connected $2$-sheeted covers of $X$?
 A: Everything you've said is correct (to prove $U_1\cap U_2$ is simply connected, just note that you can separately deformation-retract the part of it in each $\mathbb{RP}^2$ down to the basepoint to show it is contractible).  For part (c), you can use some group theory.  By the correspondence between connected covers and subgroups of $\pi_1$, the 2-sheeted covers correspond to index 2 subgroups of $\mathbb{Z}_2*\mathbb{Z}_2$.  Any index two subgroup of a group is normal, so such subgroups are exactly the kernels of epimorphisms $\mathbb{Z}_2*\mathbb{Z}_2\to\mathbb{Z}_2$.  Such epimorphisms are easy to classify by the universal property of free products: a homomorphism $\mathbb{Z}_2*\mathbb{Z}_2\to\mathbb{Z}_2$ is just a pair of homomorphisms $\mathbb{Z}_2\to\mathbb{Z}_2$, each of which is either trivial or an isomorphism.  Such a homomorphism is surjective iff at least one of the homomorphisms $\mathbb{Z}_2\to\mathbb{Z}_2$ is nontrivial.
This gives that there are three index two subgroups: the kernels of the homomorphisms $f,g,h:\mathbb{Z}_2*\mathbb{Z}_2\to\mathbb{Z}_2$ given by $f(a)=1$, $f(b)=0$, $g(a)=0$, $g(b)=1$, and $h(a)=1$, $h(b)=1$ (where $a$ and $b$ are the two generators of $\mathbb{Z}_2*\mathbb{Z}_2$).  You can then get the covering spaces as the quotients of the universal covering space by these subgroups acting as deck transformations.  More explicitly, the covering coming from $\ker(f)$ is $S^2$ (mapping to the first $\mathbb{RP}^2$) with two copies of $\mathbb{RP}^2$ (mapping to the second $\mathbb{RP}^2$) attached to it at two antipodal points.  The covering coming from $\ker(g)$ looks the same, just with the roles of the two $\mathbb{RP}^2$s in $X$ swapped.  The covering coming from $\ker(h)$ is two copies of $S^2$ attached at two antipodal points.
