I was thinking today, what is the universal cover of a torus with the "donut hole" shrunk to a point?
I am certain it must include a sphere, but that can't be enough because of the point at the center of the manifold.
Is it a wedge of two spheres perhaps?
Intuitively something like this seems correct. How should I think about these types of problems in general? Geometrically, what should I look for when searching for a universal cover?
To elaborate on @studiosus's response: Rather than thinking of a long chain of spheres, think of a long chain of sausage links.
We can think of the usual torus $T$ by starting with a cylinder, $S^1\times [0,1]$, and identifying the top and bottom circles; i.e., $T = S^1\times [0,1]/\sim$ where $(x,0)\sim (x,1)$ for every $x\in S^1$. Now, the torus with no hole, $\Sigma$, can be obtained by identifying that glued circle to a single point. I.e., we take $S^1\times [0,1]/\sim_1$, where $(x,0)\sim_1 (y,1)$ for all $x,y\in S^1$.
But let's do this a different way: First identify all the points $(x,0)\in S^1\times [0,1]$ to a point $P$ and all the points $(y,1)\in S^1\times [0,1]$ to a different point $Q$. (Topologically, this is a sphere, but we're thinking of it as a sausage link.) Now identify the points $P$ and $Q$; topologically, this is a sphere with two points identified, but we're thinking of it as attaching the ends of the sausage link to one another.
If we think of this as a sequence of two equivalence relations we're modding out by, we have $(x,0)\sim_2 (y,0)$ and $(x,1)\sim_2 (y,1)$ for all $x,y\in S^1$, and on this quotient we then identify $P$ and $Q$. Thus, we've obtained the identification space $S^1\times [0,1]/\sim_1$, so our sausage link is in fact homeomorphic to the torus with no hole, $\Sigma$.
Now, the universal cover is the infinite chain of sausage links, namely, $$S^1\times\Bbb R/\sim_4\,, \quad\text{where } (x,n)\sim_4 (y,n) \text{ for all }x,y\in S^1 \text{ and } n\in\Bbb Z.$$