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.
[Math] Universal covering space of wedge sum
algebraic-topologycovering-spacesgeneral-topology
Related Solutions
This is written for 3-sheeted covering of circle. But this is generalizable.
Take three discs and identify them along their boundary and consider the following quotient map where same coloured regions are identified. This will give you the universal covering.
It is easy to see that this space is simply connected and the cover is 3-fold.
Given a based topological space $(X,*)$, with fundamental group $G=\pi_1(X,*)$, the cover $X_H$ of $X$, with respect to a subgroup $H\subseteq G$ is defined to be:
The set of paths $\phi\colon I\to X$, such that $\phi(0)=*$, up to equivalence $\sim$. Here we say $\phi\sim \psi$ if and only if $\phi(1)=\psi(0)$ and:$$[\phi\cdot\psi^{-1}]\in H.$$ Here $\cdot$ denotes path composition. Intuitively, we are considering not just points in $X$, but a particular way of getting to them from $*$.
For connected, well behaved spaces such as finite wedges of circles or more generally finite cell complexes, we then have $X_{\{e\}}$ is the universal cover of $X$ and $X_G=X$. For the latter, note that $[\phi\cdot\psi^{-1}]$ is always in $G$, so an element of $X_G$ is determined by its endpoint.
Now let $X$ be a wedge of circles and let $*$ be the common point of the circles. Let $H$ be a normal subgroup of $G$.
Given $\phi\in X_H$ and $[g]\in\pi_1(X,*)$, we can consider $[g\cdot\phi]\in X_H$. This is well defined as if $\phi\sim \psi$ then $[\phi\cdot\psi^{-1}]=h$, for some $h\in H$ and: $$ [g\cdot\phi\cdot\psi^{-1}\cdot g^{-1}]=[g][h][g]^{-1}\in H. $$ Note we needed $H$ to be a normal subgroup for this action to be well defined.
This gives us a well defined action of $G$ on $X_H$. Note for $g\in G, x\in X_H$ we have $gx=x$ if and only if $g\in H$.
The vertices of the graph $X_H$ are precisely the translations of the basepoint $*_H\in X_H$ (the constant path $I\to X$) under the action of $G$. This is because every path ending at $*$ represents an element $g\in\pi_1(X,*)$, so equals $g*_H$.
These vertices are then in one to one correspondence with $G/H$, under the map $gH\mapsto gH*_H$. To see this note that $g*_H=f*_H$ if and only if $gf^{-1}\in H$.
We have $G=\pi_1(X,*)=*_{i=1}^n\mathbb{Z}$. Let $t_i$ be the generator of the copy of $\mathbb{Z}$ corresponding to the $i$'th circle, with edge $\vec{e_i}$. Then the edges in $X_H$ are precisely the lifts of these $\vec{e_i}$, joining vertices $gH$ and $gt_iH$.
This is precisely the Cayley graph for the group $*_{i=1}^n\mathbb{Z}/H$, with generators $t_1,\cdots,t_n$.
Conversely, for any group with $n$ generators, $g_1,\cdots, g_n$, we can let $H$ be the normal subgroup of relators in the free group on the letters $g_i$. That is words in the $g_i$ that represent the identity in G. We can identify the free group on the letters $g_i$ with $\pi_1(X,*)$, where $X$ is a wedge of $n$ circles. In particular, we let each $g_i$ represent a generator for the fundamental group of the $i$'th circle.
The Cayley graph of this group with respect to these generators is then precisely $X_H$.
Best Answer
Something fractal-like such as this? (of course, the circles are $2$-spheres.)