(a) First, we determine all the covering spaces of $S^1$. Note that $\pi_1(S^1) = \mathbb{Z}$. Since this is an abelian group, every subgroup belongs to its own trivial conjugacy class. That is, each subgroup corresponds to a unique covering space. The subgroups of $\mathbb{Z}$ are precisely $n\mathbb{Z}$ for $n \geq 0$, with $n=0$ corresponding to the trivial subgroup. For $n > 0$, the subgroup $n\mathbb{Z}$ is isomorphic to $\mathbb{Z}$. For describing covering maps, let $S^1$ be the unit circle in $\mathbb{C}$.

Corresponding to the trivial group, we have the covering space $(\mathbb{R}, p)$ where $p(x) = \exp(2 \pi i x)$.

For each $n > 0$, the covering space associated with $n\mathbb{Z}$ is $(S^1, p_n)$ where $p_n(z) = z^n$.

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(b) Now we determine all the covering spaces of $P^2$. Note $\pi_1(P_2) = \mathbb{Z}_2$. Of course, there are exactly two (non-conjugate) subgroups -- the trivial one, and the whole group. Let $P^2$ be quotient of $[0,1] \times [0,1]$ under the identification $\varphi$ corresponding to identifying the edges in pairs. Corresponding to the trivial group we have the covering space $(\mathbb{R}^2_+, p)$, where $\mathbb{R}^2_+$ denotes the first quadrant of $\mathbb{R}^2$, and the map is $p(x, y) = \varphi(x - \lfloor x \rfloor, y - \lfloor y \rfloor)$. Corresponding to the whole group we have the trivial covering space $(P^2, \text{id})$.

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(c) We now determine all the covering spaces of the annalus $X = \setbuilder{x,y \in \mathbb{R}^2}{1 \leq x^2 + y^2 \leq 4}$. However, note that this space is homeomorphic to $S^1 \times [1,4]$. The fundamental group is still $\mathbb{Z}$, since this deformation retracts to $S^1$.

Corresponding to the trivial group, we have the covering space $(\mathbb{R} \times [1,4], p)$, where the covering map is $p(x, t) = (\exp(2\pi i x), t)$.

Corresponding to $n\mathbb{Z}$, we have the covering space $(S^1 \times [1,4], p_n)$ where $p_n(z, t) = (z^n, t)$.