# Universal covering space of wedge sum

Consider the wedge sum of the unit circle and real projective plane $S^{1} \vee \mathbb{R}P^{2}$. How would one construct a universal covering space for this kind of wege sum? I've tried constructing it using an identical space to the universal covering space of $S^{1} \vee S^{2}$, which is the union of a line with infinitely many copies of $S^{2}$, except with the antipodal map, but wouldn't one need two lines that intersect with each $S^{2}$? But when I try to construct a space with two lines intersecting infinitely many $S^{2}$, I can't find a way to do it and preserve simple-connectedness.

• I think your space is a bunch of spheres arranged tangentially to a line. This is the Cayley complex of $\Bbb Z * \Bbb Z/2 = \pi_1(S^1 \vee \Bbb RP^2)$. Quotienting out by the natural action of this gives you the presentation complex of $\Bbb Z * \Bbb Z/2$, which is homotopy equivalent to $S^1 \vee \Bbb RP^2$. – Balarka Sen Mar 2 '15 at 19:47
• So the space is equivalent to the universal covering of $S^{1} \vee S^{2}$? – user211685 Mar 2 '15 at 19:50

You can construct the universal cover of $\mathbb{R}P^2\vee S^1$ in the following way: you can first construct a double cover, i.e. $S^1\vee S^2\vee S^1$. The universal cover of this space can be constructed by considering the well known universal cover of $S^1\vee S^1$, but you have to consider that now you have an extra $S^2$. You can see it as an infinite layer Cayley graph in which between every layer there is an $S^2$.

This space is homotopically equivalent to the Cayley graph with an $S^2$ attached to each node, which is indeed equivalent to an infinite wedge sum of $S^2$'s.

• Would the construction with a real line, and an $S^{2}$ attached at integer points on it work? – user211685 Mar 2 '15 at 20:08
• @user211685 Sure. That's homotopy equivalent to $\bigvee S^2$. – Balarka Sen Mar 2 '15 at 20:09
• @user211685 No, because some neighborhood of the "north pole" (not on the line) should be homeomorphic to some neighborhood of the base point in $S^1\vee RP^2$ – Peter Franek Mar 2 '15 at 20:10
• It would not be homeomorphic to the universal cover you are looking for, but indeed it would be homotopically equivalent. – Dario Mar 2 '15 at 20:12
• So basically, only an infinite layer cayley graph with $S^{2}$ such as in the answer below, would suffice as a construction? – user211685 Mar 2 '15 at 20:16

Something fractal-like such as this? (of course, the circles are $2$-spheres.)

• I have no idea why it should be so. What's the covering map you have in mind? – Balarka Sen Mar 2 '15 at 20:01
• @BalarkaSen The spheres (circle in the picture) are projected to the $RP^2$. Two oposite points on the sphere are mapped to the base point in the wedge sum.. Or is it a nonsense? – Peter Franek Mar 2 '15 at 20:03