I have the following question from a past exam about basic algebraic topology:
Let $p:S^1 \to \mathbb{R}\mathbb{P}^1$ be the obvious covering map. For $n\geq 2$, show that a continuous map $f:\mathbb{R}\mathbb{P}^n \to \mathbb{R}\mathbb{P}^1$ always lifts to a continuous map $\tilde{f}:\mathbb{R}\mathbb{P}^n \to S^1$ with $p \circ \tilde{f} = f$.
Deduce that, for all $n \geq 2$, there is no continuous map $F:S^n \to S^1$ with $F(x) = -F(-x)$ for all $x$.
I have no trouble with the first part, and am able to make some headway into the second, but I need help getting the result. The examiner's report says that one is supposed to use uniqueness of lifts to finish the argument, and I don't see how to apply it.
My Attempt
The first part is straightforward. Note that $\mathbb{R}\mathbb{P}^n$ has fundamental group $\mathbb{Z}/2\mathbb{Z}$. The homomorphism $f_*:\pi_1(\mathbb{R}\mathbb{P}^n) \to \pi_1(\mathbb{R}\mathbb{P^1})\cong \mathbb{Z}$ thus maps the nontrivial element of $\pi_1(\mathbb{R}\mathbb{P}^n)$ to an element $n$ of $\mathbb{Z}$ with $2n = 0$, hence $n = 0$, so $f_*\pi_1(\mathbb{R}\mathbb{P}^n)$ is trivial.
In particular $f_*\pi_1(\mathbb{R}\mathbb{P}^n, y) \subseteq p_*\pi_1(S^1, \tilde b)$, where we have chosen basepoints $y, \tilde b, b$ for our spaces such that $f(y) = b, p(\tilde b) = b$. Now, since $\mathbb{R}\mathbb{P}^n$ is path-connected and locally path-connected, by a standard theorem referred to in my course as existence of lifts, we have that $f$ admits a lift $\tilde f:(\mathbb{R}\mathbb{P}^n, y)\to (S^1,\tilde{b})$.
Let $q:S^n\to\mathbb{R}\mathbb{P}^n$ be the obvious two-sheeted covering. Suppose that a continuous odd $F:S^n \to S^1$ exists. The condition that $F$ is odd means that for $x, y \in S^n$, we have $$q(x) = q(y) \implies y = \pm x \implies F(y) = \pm F(x) \implies p(F(x)) = p(F(y))$$ so there is a well-defined map $f:\mathbb{R}\mathbb{P}^n \to \mathbb{R}\mathbb{P}^1$ with $f(q(x)) = p(F(x))$ for $x \in S^n$. So the following diagram commutes:
Now, by our result from the first part of the question, $f$ actually has a lift $\tilde f:\mathbb{R}\mathbb{P}^n \to S^1$. From here I'm not sure how to proceed. The hint in the examiner's report says I should use uniqueness of lifts, but here we only have one lift, so I don't see a way to do this. I guess we want to construct some distinct lift that agrees with $\tilde f$ at some point, but I haven't been able to find such a construction.
You have made a choice (which is always a hazard for uniqueness) when you chose $+F(x)$ from $\pm F(x)$ to construct your $f$. Let's call that map $f^+$ to record that choice. There is also an $f^-$ constructed from $-F(x)$. This gives two lower maps in your commutative diagram (parallel to the two upper maps) and the two are not the same. Consequently, there are two distinct commuting paths in the diagram, but they do not "mix" -- whatever map you pick on the upper side you must pick the matching map on the lower side -- from $\mathbb{R}P^n$ to $S^1$, giving distinct lifts $\widetilde{f^+}$ and $\widetilde{f^-}$, contradicting uniqueness of lifts.
A better diagram would be $\require{AMScd}$ \begin{CD} S^1 @<{-F}<< S^n @>{+F}>> S^1\\ @VVV @VVV @VVV \\ \Bbb{R}P^1 @<{f^-}<< \Bbb{R}P^n @>{f^+}>> \Bbb{R}P^1 \end{CD}
This diagram commutes. However if we add the identity map between the pair of $S^1$s or between the pair of $\Bbb{R}P^1$s or between both pairs, the resulting diagram does not commute, explicitly demonstrating a failure of uniqueness of lifts.