Let $\sim$ be the equivalence relation $$a\sim b\iff a=b\text{ or }a=-b,$$ for $a,b$ on the unit sphere $S^2$.
Let $Q$ be the quotient space.

  1. How do I show that the quotient map is a covering map?
  2. Let $q\in Q$. How do I show that the fundamental group $\pi_1(Q,q)$ has order $2$?

What I know:

  1. So $Q=S^2\backslash\sim$ and the quotient map is $f:S^2\rightarrow Q$ with $f(x)=[x]$. The quotient topology is $T_Q=\{U\subset Q\}$ such that $f^{-1}U$ is open in $S^2$.
    Further a covering space of $S^2$ is a pair $(Y,p)$ with $Y$ a topological space and $p$ a continuous surjective map such that:

    • $S^2$ has an open cover $U$ such that for all $u\in U$ we have that $p^{-1}u\subset Y$ is a disjunct union of open sets $v\subset Y$ for which $p|_v:v\rightarrow u$ is a homeomorphism.
      Then $p$ is a covering map.
      It is obvious that $f$ is surjective and continuous, but how do I prove the characteristic above?
  2. All the closed paths $\gamma:[0,1]\rightarrow Q$ with $\gamma(0)=\gamma(1)=q$ are in the set $P(Q,q)$. Path homotopy is the equivalence relation which we use to find the quotient $\pi_1(Q,q)$ of $P(Q,q)$.
    Does that mean that the question wants me to show there are only two equivalence classes in $\pi_1(Q,q)$? And how would I do that?

  • $\begingroup$ As an aside, your $Q$ is known as the real projective plane. $\endgroup$ May 3, 2016 at 13:02
  • $\begingroup$ @HenningMakholm Yes, my question stated that, but I did not think it was necessary to add that here. Should I have? $\endgroup$
    – user335050
    May 3, 2016 at 13:12
  • $\begingroup$ @HenningMakholm Could you maybe help with the questions? $\endgroup$
    – user335050
    May 3, 2016 at 13:14
  • $\begingroup$ No, unfortunately I'm not really proficient enough with these matters to do more than shout aproposes from the sidelines. $\endgroup$ May 3, 2016 at 13:15
  • $\begingroup$ @HenningMakholm Oh well, thanks anyway $\endgroup$
    – user335050
    May 3, 2016 at 13:17

1 Answer 1


It looks like you've got a few things backwards in question 1. You're not looking for a covering space of $S^2$ - there are no such things other than $S^2$ itself, since it's simply connected!

Instead, you're looking for a covering space of $Q$. In fact, you're looking to show that $S^2$ is a covering space of $Q$. (The covering map should be $f$, so using your notation you want $(S^2, f)$ to be a covering space of $Q$.)

You've also got the covering space condition a bit mucked up. You want an open cover $\{U_i\}$ of $Q$ - this is going to be a set of open sets in $Q$ - so that $f^{-1}(U_i)$ is a disjoint union of open sets $\coprod_j V_{ij}$, and $f|_{V_{ij}}: V_{ij} \to U_i$ is a homeomorphism.

Now your covering map is two-to-one (since it identifies two points in $S^2$ to each point in $Q$), so $f^{-1}(U_i)$ is going to be a disjoint union of two open sets $V_{i,1} \coprod V_{i,2}$.

How can this fail to happen? Well, let's suppose one of our open sets $U_i \subset Q$ is really big - maybe it's all of $Q$. Then when we lift it to $S^2$, we just get all of $S^2$. Since $S^2$ is connected, this isn't a disjoint union of two open sets!! So we need all our $U_i$ to be really small. It helps here if you think about $\mathbb{RP}^2$ as being a half-sphere with its boundary identified in a certain way.

Essentially, you want to cover $S^2$ by pairs of disjoint open sets $V_i \coprod -V_i$ (since your covering map will identify $v \in V_i$ and $-v \in V_i$ in $Q$. And if an open subset $V \subset S^2$ does not intersect $-V$, it has to be pretty small. Then showing what you want is pretty easy: the inverse image of $[V_i] = \{[v]| v \in V_i\} \subset Q$ is clearly $V_i \coprod -V_i$, and both $V_i$ and $-V_i$ are easily seen to be homeomorphic to $[V_i]$.

For question 2, things are a bit easier. Fix a point $x_0 \in S^2$. The trick here is to realize that any closed path in $Q$ starting at $[x_0]$, when lifted to $S^2$ starting at $x_0$ (and think hard about why you can uniquely lift paths!) must either start and end at $x_0$, or start at $x_0$ and end at $-x_0$.

If your path starts and ends at $x_0$, then remember that $S^2$ is simply connected and unpack that definition. If your path starts at $x_0$ and ends at $-x_0$, then you've found a path that can't be null-homotopic in $Q$. If you lift twice this path, however, it will start at $x_0$, pass through $-x_0$, and end again at $x_0$, and will thus be null-homotopic again! So now all you need to do to finish is to show that all paths starting at $x_0$ and ending at $-x_0$ are homotopic, and this is straightforward - any path starting at $x_0$ and ending at $-x_0$ can be written as a path from $x_0$ to itself, followed by a fixed path $x_0$ to $-x_0$, so they're all homotopic!

On the whole, it seems like you need to remember that when thinking about covering spaces, it's usually helpful to do all your work upstairs in the covering space where life is simpler. $Q$ is a nasty space to get a hang of at a down-and-dirty level; just lift things up to $S^2$ where life is nice and easy!

  • $\begingroup$ Thank you very much! You were right, I did get my definitions a bit wrong there. This is a really nicely written answer by the way :) $\endgroup$
    – user335050
    May 3, 2016 at 15:40
  • $\begingroup$ I dont fully understand the homotopy of all paths that begin at $x_0$ and end at $-x_0$. What do you mean by lifting the path 'twice'? And what do you mean by writing a path from $x_0$ to itself followed by a fixed path from $x_0$ to $-x_0$? $\endgroup$
    – user335050
    May 9, 2016 at 11:12

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