Denote for every $n \in \Bbb{Z}, n\geq 1$ the Euclidean topology of $\Bbb{R}^n$ as $\mathcal{T}_E^n$.
From my Algebraic topology lesssons, I have the following excercise:
Consider $C:=\{(x,y,z) \in \Bbb{R}^3 : x^2+y^2=1 \}$ as a subset of the Euclidean three-dimensional real space $(C, \mathcal{T}^3_E |_{C})$.
- Compute an universal convering space $(\tilde{C}, p)$ for $C$.
- Describe topologically $\tilde{C}$ and the aplicaction $p: \tilde{C} \rightarrow C$.
If I am not wrong, in order to do this we can view $C$ as $$(C, \mathcal{T}_E^3)=(\Bbb{S}^1 \times \Bbb{R}, \mathcal{T}_E^2|_{\Bbb{S}^1} \times \mathcal{T}_E^1)$$ Since $(\Bbb{R}, \mathcal{T}_E^1)$ is an universal covering space of $(\Bbb{S}^1, \mathcal{T}_E^2 |_{\Bbb{S}^1})$ vía $r: \Bbb{R} \rightarrow \Bbb{S}^1, r(t)=(\cos (t),\sin (t))$, we can take $\tilde{C} := \Bbb{R} \times \Bbb{R} (= \Bbb{R}^2)$ and $p: \tilde{C} \rightarrow C, p(t,s)= (\cos (t), \sin (t), s)$ and 1. is finished.
For 2., the only thing I think that I would say is that $\mathcal{T}_E^1 \times \mathcal{T}_E^1 = \mathcal{T}_E^2$ and therefore $(\Bbb{R} \times \Bbb{R}, \mathcal{T}_E^1 \times \mathcal{T}_E^1) = (\Bbb{R}^2, \mathcal{T}_E^2)$ so $\tilde{C}=(\Bbb{R}^2, \mathcal{T}_E^2)$.
Is this fine? Since it is that short I have doubts, so any confirmation or correction would be appreciated :)