We know that topologically speaking, $$\text{SO}(3) \simeq \frac{D^3}{\sim_A}\simeq \frac{\mathbb{S}^3}{\sim_A}\simeq \mathbb{RP}^3 $$ where $\sim_A$ is the equivalence relation identifying antipodal points. I have spent quite some time understanding this diffeomorphism, and the derivation now makes sense to me. Using it, however, to find the answer to a question I've been wondering about is trickier. Suppose we consider the 'north pole' of $\mathbb{S}^3$, which is defined in Cartesian coordinates as $(0,0,0,1)$. On $\mathbb{RP^3}$, this is identified with the point $(0,0,0,-1)$, i.e., the 'south pole' of the 3-sphere. Suppose I want to find the rotation corresponding to this point of $\mathbb{RP}^3$ on SO(3). That is, to find the rotation ('point') on $SO(3)$ that corresponds to the single element of $\mathbb{RP}^3$ identifying the north and south poles on the three-sphere.
Intuitively, I expect the rotation to be by $\pi$ radians. Identifying the axis is where the trouble comes in. I am not sure if quaternions are the most direct approach to this. Any tips would be appreciated.
I do believe quaternions are the most direct approach to it. That is to say: to computing the rotation that corresponds to a given point on $\mathbb{S}^3/\sim_A$. Showing that this correspondence is bijective is a different story, but you said you already understood that part.
You know that elements of $SO(3)$ are rotations of $\mathbb{R}^3$. So in the simplest form an answer to your question would be, given an element $\overline{q}$ in $\mathbb{RP}^3$, a recipe that tells you for each $x \in \mathbb{R}^3$ to what point in $\mathbb{R}^3$ the point $x$ is mapped by the rotation. (A more complicated form of answer would be giving the axis and angle of rotation, I come back to that later.)
Now in this 'where does $x$ go?'-form the answer is really simple.
Step 1: pick a representative $q \in \mathbb{S}^3$ of the equivalence class $\overline{q}$.
Step 2: View $q$ as a quaternion, so if you would have ordinarily written it as $(0, 0, 0, 1)$ it now becomes $0 + 0i + 0j + 1k = k$
Step 3: View $x$ as a pure quaternion, i.e. a quaternion with no real part, so if $x$ would normally written $(x_1, x_2, x_3)$ it is now written $x_1i + x_2j + x_3k$.
Step 4: compute $qxq^{-1}$. This is also a pure quaternion, and more importantly: this is your answer!!
Step 5 (optional): check that this does not depend on the choice of representative $q$. This is not so hard: the only other element in the equivalence class is $-q$ and since $(-q)^{-1} = -q^{-1}$ we find that using $-q$ instead of $q$ would have led to $qxq^{-1}$ as well since the minus signs in $(-q)x(-q)^{-1}$ cancel eachother out.
So yeah, the answer is just the rotation $x \mapsto qxq^{-1}$, which is especially easy to compute because for $q = a + bi + cj + dk \in \mathbb{S}^3$ we have that $q^{-1} = a - bi - cj - dk$.
(The general principle here is that $q^{-1} = (a - bi - cj - dk)/(a^2 + b^2 + c^2 + d^2)$, but since $q$ lives in $\mathbb{S}^3$ we have that $(a^2 + b^2 + c^2 + d^2) = 1$)
(Obvious from this formulation it is not clear that this maps $\mathbb{R}^3$ into $\mathbb{R}^3$, i.e. that the resulting quaternions have no real part, let alone that it is a rotation, but when the question is purely 'how to compute the map' this is the answer. We can talk about the proof in a different question/answer on this site. (There probably is already a question/answer with this proof on the site somewhere)).
Now for the axis of rotation. We can write $q = a + y$ with $a \in \mathbb{R}$ and $y$ a pure quaternion. My claim is that $y$ (and hence all its multiples) are mapped to itself and hence the line through $y$ and $-y$ in $\mathbb{R}^3 = \mathbb{R}i + \mathbb{R}j + \mathbb{R}k$ is the rotation axis.
Proof: $qyq^{-1} = (a + y)y(a - y)$. Since $y$ is a pure quaternion we have that $y^2$ is a negative real number. Moreover, since the length of $q$ equals 1, that number moreover satisfies $a^2 - y^2 = 1$. Now expanding out $qyq^{-1} = (a + y)y(a - y)$ using this knowledge we find that the real part of the outcome equals $ay^2 - ay^2 = 0$ (as it should to keep the outcome inside the pure quaterions) and that the 'pure quaternion'-part equals $a^2y - y^3 = (a^2 - y^2)y = y$, as predicted.
Getting the angle of rotation is a bit more tricky, maybe you can figure it out for yourself now that you know around what axis we are rotating.