Universal covering space of wedge sum

Question

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.

Answer 1

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

Answer 2

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.