Find the universal covering of $\mathbb{R}P^2\vee\mathbb{R}P^2$

Question

Find the universal covering of $\mathbb{R}P^2\vee\mathbb{R}P^2$

I know that $\mathbb{R}P^2\cong \mathbb{S}^2/\sim$ for antipodal action and that $\pi_1(\mathbb{R}P^2)=\mathbb{Z}_2$, with which $\pi_1(\mathbb{R}P^2\vee\mathbb{R}P^2)=\mathbb{Z}_2*\mathbb{Z}_2$, so the universal covering of $\mathbb{R}P^2\vee\mathbb{R}P^2$ corresponds to the trivial subgroup of $\mathbb{Z}_2*\mathbb{Z}_2$, I know that $\mathbb{R}P^2\vee\mathbb{R}P^2$ has infinite covering spaces like for example: $\mathbb{R}P^2\vee\mathbb{R}P^2, \mathbb{S}^2\vee\mathbb{S}^2, \mathbb{S}^2\vee\mathbb{S}^2\vee\mathbb{S}^2$ and besides I know that an infinite chain of $\mathbb{S}^2$ glued tangent is the universal covering space but I do not know how to prove this formally in a precise way, I know that the universal covering space fulfills that it is covering space and that it is also simply connected, but my doubt is that why the space Universal covering is not $\mathbb{S}^2\vee\mathbb{S}^2$? Is it simply connected? Why is an infinite chain of spheres stuck tangentially simply connected?

Answer 1

Well, because covers are required to have locally homeomorphic fibers. What would be the preimage of the single point attaching $\mathbb RP^n \vee \mathbb RP^n$?

A "formal" and "precise" way to prove that the infinite chain is the correct universal cover, is to use describe the covering map, and state that since this space is simply connected, it is universal.

---

Perhaps worth mentioning: there is a cover using finitely many spheres, but it is not simply connected. The basic way to construct it, is to take an even number of spheres, and form a "ring" of them. The restriction that they be even is best understood (at least in my eyes) from the interpretation of $\pi_1(\mathbb RP^2 \vee \mathbb RP^2)=\mathbb Z_2*\mathbb Z_2=\langle a,b\mid a^2,b^2\rangle$. In this case, the $1$ skeleton of the universal cover is a graph alternating in $a$ and $b$. So, you need an even number of them to correspond to the subgroup $(ab)^n$, with $n>1$. The fundamental group of this space is $\mathbb Z$.

Answer 2

So first you want to know why an infinite string of spheres going off to infinity in both directions is the universal cover of $\Bbb RP^2\vee\Bbb RP^2$.

Let $\ast$ be the point of $\Bbb RP^2$ which comes from the points $N$ and $S$ of $S^2$ under the antipodal action, which are the north and south poles of $S^2$. Let $X$ be the space which is the disjoint union of two copies of $S^2$, $S^1\times\{0\}$ and $S^1\times\{0\}$, gluing their north poles with one another and their south poles with one another. Then we can consider the function $p:X\rightarrow \Bbb RP^2\vee\Bbb RP^2$ which on the first copy of $S^2$ of $X$ is the antipodal map onto the first copy of $\Bbb RP^2$ in $\Bbb RP^2\vee\Bbb RP^2$ and on the second copy of $S^2$ of $X$ onto the second copy of $\Bbb RP^2$ in $\Bbb RP^2\vee\Bbb RP^2$.

Then as the antipodal is a covering map, prove that $p$ is also a covering map; notice that the definition clearly holds for all points of $\Bbb RP^2\vee\Bbb RP^2$ different from $\ast$, so you have to see why it holds for $\ast$.

Now consider $\Bbb Z$ many copies of $S^2$, $S^2\times\{n\}, n\in\Bbb Z$, and let $X_\infty$ be the quotient obtained by gluing for each $n$ the north pole of $S^2\times\{n\}$ with the north pole of $S^2\times\{n+1\}$ and the south pole of $S^2\times\{n\}$ with the south pole of $S^2\times\{n-1\}$ if $n$ is even, and the north pole of $S^2\times\{n\}$ with the north pole of $S^2\times\{n-1\}$ and the south pole of $S^2\times\{n\}$ with the south pole of $S^2\times\{n+1\}$ if $n$ is odd.

Now consider the map $q:X\rightarrow X_\infty$ such that for each $n$, $q$ restricted to $S^2\times\{n\}$ is the identity map $S^2\times\{n\}\rightarrow S^2\times\{i\}$, where $i=0,1$ is such that $n\equiv i\mod 2$. Then it is easy to check $q$ is a covering map.

Thus $X_\infty$ is a covering space of $\Bbb RP^2\vee\Bbb RP^2$.

You also want to know why $X_\infty$ is simply connected. As $S^2$ is simply connected, it follows from Van-Kampen theorem and induction that any finite substring of $X_\infty$ is simply connected. Now any loop $\gamma$ in $X_\infty$ is contained in a finite substring of $X_\infty$, as $\gamma$ is compact and thus it is homotopic to the constant loop.

Finally, you ask why $S^2\vee S^2$ is not the universal cover of $\Bbb RP^2\vee\Bbb RP^2$. Well, the thing is that $S^2\vee S^2$ is not even a covering space of $\Bbb RP^2\vee\Bbb RP^2$. This is because $\Bbb RP^2\vee\Bbb RP^2$ is almost a surface except for the singularity $\ast$; which is a point having no open nhood homeomorphic to $\Bbb R^2$, so if $S^2\vee S^2$ were a covering space of $\Bbb RP^2\vee\Bbb RP^2$ it should have at leat two of this singularities, but $S^2\vee S^2$ has only one such singularity.