Here is the problem statement:
Let $p:X\to S^1$, $q:Y\to S^1$ be covering maps, and $s_0=1\in S^1\subseteq \mathbb C$ is a basepoint for the unit circle. Let $M$ be the set of continuous maps $f:X\to Y$ such that $qf = p$.
(a) Suppose $X = \mathbb R$ and $p(t) = e^{2\pi it}$. Describe and prove bijections $M\xrightarrow{\sim} q^{-1}(s_0)$. (We do not assume that $Y$ is path-connected.)
(b) Suppose $X = S^1$ and $p(z) = z^2$. Give an example of a covering map $q:Y\to S^1$ such that $M$ is empty.
I have a general idea for part (a), but I would like some help proofreading (and/or formalizing) it: note that $X = \mathbb R$ is the universal covering for $(S^1,s_0)$ in this case, and note that we require a choice of basepoint on the universal covering to work on a correspondence. Let us fix a basepoint $x_0\in X$, i.e., $x_0\in p^{-1}(s_0)$ is some point in the fiber of $1$ over $p$. If we pick an arbitrary basepoint $y_0\in q^{-1}(s_0)$, then both $p$ and $q$ are pointed maps, and by the universal property (as the initial object of the category of covering spaces), there exists a unique continuous mapping $f:(X,x_0)\to (Y,y_0)$ that makes the diagram commute. (I guess I am assuming $Y$ to be connected?) This is the well-defined assignment we want from a choice of basepoint $y_0\in q^{-1}(s_0)$ to a covering map $f:X\to Y$. As for the backwards direction, since we fixed $x_0\in X$ beforehand, then a continuous map $f:X\to Y$ such that $qf = p$ sends $x_0$ to some point $y_0\in Y$. In particular, $y_0$ lands in the preimage of $s_0\in S^1$ by the commutativity of the diagram. These assignments give the bijection we want.
I have not made any progress on part (b) though...? (I tried applying Borsuk-Ulam theorem but it did not work.) Any help would be appreciated!
Update: here is my solution to part (b) following the hints in the comments.
Let $Y = \mathbb R$ and let $q:Y\to S^1$ be a covering map. Suppose $M\neq \varnothing$, so we take $f:X\to Y$ to be a continuous map that makes the diagram commute. Note that $X = S^1$ is path-connected and locally path-connected. By the lifting criterion, we know such $f:X\to Y$ exists if and only if $p_*(\pi_1(X,x_0))\subseteq q_*(\pi_1(Y,y_0))$. Plugging things in, we have $p_*(\mathbb Z)\subseteq q_*(0) = 0$, so $p_*(\mathbb Z) = 0$, namely $p_*$ is the zero map. But we know $p_*:\pi_1(X,x_0)\to \pi_1(S^1,s_0)$ is induced by $p:X\to S^1$ given by $p(z) = z^2$, so $p_*$ is given by multiplication by $2$, contradiction. Therefore, such $f$ does not exist.
Your ideas are correct. I suggest some "streamlining".
(a)
Let us first consider any map $p : X \to S^1$ defined on a connected space $X$. The continuous maps $f : X \to Y$ with $q f = p$ are nothing else than the lifts of $p : X \to S^1$.
Choose a basepoint $x_0 \in X$. Since $X$ is connected, two lifts $f, f'$ of $p$ agree if and only if they agree at the point $x_0$ (i.e $f(x_0) = f'(x_0)$). For each lift $f$ of $p$ we have $f(x_0) \in q^{-1}(p(x_0))$. Thus we get an injection $$\lambda : M \to q^{-1}(p(x_0)), \lambda(f) = f(x_0) .$$ Our task is to determine the image of $\lambda$. The well-known lifting theorem gives the answer for any path-connected and locally path-connected space $X$:
Let $y_0 \in q^{-1}(p(x_0))$. Then a lift $f : X \to Y$ with $f(x_0) = y_0$ exists if and only if $p_*(\pi_1(X,x_0)) \subset q_*(\pi_1(Y,y_0))$.
In other words, $y_0 \in \lambda(M)$ if and only if $p_*(\pi_1(X,x_0)) \subset q_*(\pi_1(Y,y_0))$.
Therefore $\lambda$ is always a bijection if $X$ is simply connected and locally path-connected.
This applies to $X = \mathbb R$ (for any map $p$). Taking $x_0 = 0$ as a basepoint for $\mathbb R$ and $p(t) = e^{2\pi i t}$ we get $p(0) = s_0$ and see that $$\lambda : M \to q^{-1}(s_0)$$ is a bijection.
(b)
Let $p_n : S^1 \to S^1, p_n(z) = z^n$. We get $(p_n)_*(\pi_1(S^1,s_0)) = n\mathbb Z$. Thus for $p = p_2$ the coverings $p_n$ with $n \ne 2$ and the covering $e : \mathbb R \to S^1, e(t) = e^{2\pi i t}$, provide examples for which $M$ is empty.