# Can the real projective plane be considered as a covering space of a closed disk?

I was reading that covering spaces are a topological generalization of Riemann surfaces.

Thinking about the Riemann surface defined by the complex logarithm, smooth movement along that surface can cause discontinuous jumps when projected down onto the complex plane (i.e. over a branch cut).

A similar phenomenon seems to occur when projecting the real projective plane down onto the unit hemisphere (which is homeomorphic to a closed disk) -- see around 0:55 in this animation on Youtube: https://www.youtube.com/watch?v=x2SZSfYYSc8

Question: There is a clear visual analogy here -- is there a rigorously defined topological one as well? Can the real projective plane be construed as a covering space for the unit hemisphere?

• The depiction of the real projective plane using the closed unit disk is an implied mapping $D^2\to\mathbb{RP}^2$ not the other way around. So you seem to have it backwards. – arctic tern Jul 30 '16 at 1:07
• The closed disk is simply connected, so it has no nontrivial covering spaces. In the other direction, the only nontrivial (connected) covering space of the real projective plane is the sphere $S^2$. – Qiaochu Yuan Jul 30 '16 at 5:18
• In some sense it is: There is an involution s of the projective plane such that the quotient $RP^2/s$ is homeomorphic to the closed disk. The difference with the usual covering spaces is that this involution has some fixed points. – Moishe Kohan Jul 30 '16 at 15:34

When depicting $\mathbb{RP}^2$ using the unit disk $D^2$, there is an implied mapping $D^2\to\mathbb{RP}^2$. But this is not a covering map. Some points in $\mathbb{RP}^2$ come from two different points of $D^2$ (namely antipodal points on the boundary circle $S^1$) whereas other points in $\mathbb{RP}^2$ come from a single point in $D^2$ (namely interior points). Because of this mismatch in the size of fibers ("fiber" meaning the preimage of a singleton set), it cannot be a covering map.
You're right that their is a faint similarity between the two situations though. As one traverses the branch cut of the logarithm, the output jumps by $2\pi i$. And as one traverses across the image of $S^2$ within the projective plane $\mathbb{RP}^2$, the inverse image jumps from one side of $D^2$ to the other. In both cases this has to do with one space being a quotient space of the other.
Given any topological space $X$ and equivalence relation $\sim$ defined on it (i.e. a binary relation which is reflexive, symmetric and transitive just like equality is), there is a quotient space $X/\sim$ and a quotient mapping $X\to X/\sim$. The effect of this is "identifying" points which are related by $\sim$, for instance on $D^2$ we can define $x\sim y$ by: either $x=y$ in the interior of $D^2$, or $x=\pm y$ on the boundary $S^2$. Then $D^2/\sim$ is basically an avatar for $\mathbb{RP}^2$.
Similarly, the Riemann surface associated to the logarithm function is an infinite staircase, where each "stair" is a ray (homeomorphic to $\mathbb{R}$) and the stairs extend infinitely high and low (so along an axis $\mathbb{R}$). This Riemann surface is just $\mathbb{C}$ itself. We have an equivalence relation on it where $x\sim y$ whenever $x$ and $y$ are on staircases that sit exactly one over the other (by some integer multiple of $2\pi i$, heuristically speaking). Then the quotient space is just the complex plane minus $0$. The quotient map $\mathbb{C}\to\mathbb{C}^\times$ is in fact just the exponential.
Basically, any surjective continuous function between topological spaces can be interpreted as a quotient map (where $\sim$ is defined by $x\sim y$ whenever $f(x)=f(y)$). But covering spaces are very special cases of continuous maps: the size of the fibers are constant (although this is not sufficient, since maps can still "tangle" things incorrectly - instead it needs to apply to all sufficiently small neighborhoods of all points).