# This covering map is homeomorphism

Suppose that $$f:\mathbf RP^2\rightarrow X$$ is a covering map and $$X$$ is a CW-complex. Show that $$f$$ is homoemorphism.

We know covering map is continuous and onto,so we should show that $$f$$ is one-to-one and the inverse exists(which by the definition of covering map is trivial too) and is continuous,but i really don't have any idea to show this continuity and one-to-one property of $$f$$ . Could you help me with this problem?

Thanks in advance

• All you need to get is 1-1ness, since $\mathbb{R}P^2$ is compact and $X$ is Hausdorff. – Randall Jan 22 at 17:49
• @Randall : or because the inverse (if it exists) is locally continuous, because $f$ is a covering map. – Max Jan 22 at 18:26
• @Max i think locally continuous gives us locally homeomorphism not homemorphism,am i right? – pershina olad Jan 22 at 20:59
• Yes but a bijection which is a local homeomorphism is a homeomorphism – Max Jan 22 at 21:15
• o! I didnt know that.Thank you for your answer and good comments @Max – pershina olad Jan 25 at 6:53

## 2 Answers

Here's a sketch of a solution :

1) If a group acts freely on the sphere $$S^2$$, then it's $$1$$ or $$\mathbb{Z/2Z}$$

To prove this, note that $$\pi_2(S^2)\simeq \mathbb{Z}$$, so if $$G$$ acts freely on $$S^2$$ one can define, for $$g\in G$$, $$d(g)$$ to be $$1$$ or $$-1$$ depending on whether $$g_* : \pi_2(S^2)\to \pi_2(S^2)$$ is $$id$$ or $$-id$$. $$d$$ is clearly a morphism.

Now if $$g\in G$$ has no fixed points, then its action on $$S^2$$ is homotopic to $$-id : S^2\to S^2$$ by $$H(t,x) = \frac{t g\cdot x - (1-t)x}{||t g\cdot x - (1-t)x||}$$, this being well defined because by assumption, $$t g\cdot x - (1-t)x$$ cannot vanish; so $$g_* = (-id)_* = (-id)^3 = -id$$, so $$d(g) = -1$$.

Thus if the action is free, and $$g\neq e$$, then $$d(g)\neq 1$$: $$d$$ is injective, thus providing the result.

2) If $$p:Y\to X$$ is a covering map, $$X$$ is $$T_1$$ and $$Y$$ is compact, then it's a finite sheeted covering.

This is an easy topology exercise.

3) If $$q:Z\to Y, p:Y\to X$$ are covering maps and $$p$$ is a finite sheeted covering, then $$p\circ q$$ is also a covering.

This is again, an easy topology exercise, you just have to play with the definitions of covering and remember that a finite intersection of open sets is open.

4) If $$p:Y\to X$$ is a covering map, $$X$$ is a nice space (connected, locally path connected, semi-locally simply-connected) and $$Y$$ is simply connected, then this covering map is equivalent to a canonical projection $$\pi : Y\to Y/G$$ for some group $$G$$ acting freely and properly discontinuously on $$Y$$.

This is standard covering theory. The group $$G$$ will be $$\pi_1(X,x)$$ for some $$x\in X$$, and the action will be the classical one, i.e. for some fixed $$y\in p^{-1}(x)$$; for all $$g\in G$$, lift $$g$$ uniquely as a path $$\gamma$$ starting from $$y$$, then there is a unique deck transformation of $$Y$$ that sends $$y$$ to $$\gamma(1)$$ : this deck transformation is the action of $$g$$.

Put $$G=\pi_1(X,x)$$ for some fixed $$x\in X$$. Then $$p:Y\to X$$ is clearly $$G$$-invariant and so factors uniquely through some $$f: Y/G\to X$$. Now $$f$$ is clearly a continuous bijection, it takes a bit more work to show that it is a homeomorphism, but that's essentially because $$p$$ is a local homeomorphism.

But it doesn't matter because we don't need 4) altogether, we only need it when $$Y$$ is compact, and $$X$$ $$T_2$$ in which case, continuous bijection implies homeomorphism, so there's no additional work to do.

5) To patch things up : By 2), your $$f$$ is a finite sheeted covering. Thus by 3), the composition $$S^2\to \mathbf{R}P^2\to X$$ is also a covering; and $$S^2$$ is simply connected and compact, and $$X$$ is a nice space because it is a connected CW-complex, so $$X \simeq S^2/G$$ for $$G=\pi_1(X,x)$$ acting freely on $$S^2$$. But by 1), this implies $$\pi_1(X,x) = 1$$ or $$\mathbb{Z/2Z}$$. It can't be one, because $$f_* : \pi_1(\mathbf{R}P^2)\to \pi_1(X)$$ is injective, so it must be $$\mathbb{Z/2Z}$$.

But then $$f_*$$ is an isomorphism, and this is again standard covering theory : this implies that $$f$$ is a $$1$$-sheeted covering, i.e. a homeomorphism.

• Clearly your proof applies to any even dimensional projective space. Do you know if there is a counterexample of the claim if we replace $\mathbb{R}P^2$ by an odd dimensional projective space? Maybe if $X$ is some Lens space? – Lukas Jan 22 at 19:47
• @Lukas : you are right about the even dimensions. About odd dimensions, there's at least the stupid example of $\mathbf{R}P^1 \simeq S^1$, but the covering $S^1\to \mathbf{R}P^1$ is not itself a homeomorphism (it's a $2$-sheeted covering). For higher odd dimensional projective space, it would amount to finding a finite group $G$ acting freely on $S^{2n+1}$, with an action strictly containing $\mathbb{Z/2Z}$. I'm not sure but I don't see any reason why a quotient of a lens space couldn't work : (1/2) – Max Jan 22 at 21:25
• Say if you take a lens space $L(p;q)$ with $p,q$ coprime and odd, then $S^3$ should inherit a $\mathbb{Z/p\times Z/2}$ action of the form $(k,0)\cdot (z_1,z_2) = (e^{\frac{2ik\pi}{p}}z_1 , e^{\frac{2ikq\pi}{p}}z_2)$ and $(k,1)\cdot (z_1,z_2) = (-e^{\frac{2ik\pi}{p}}z_1 , -e^{\frac{2ikq\pi}{p}}z_2)$, and this should be free (that's where the choice of $p,q$ odd intervenes); so that $\mathbf{R}P^3$ inherits a free $\mathbb{Z/p}$-action; and a free $\mathbb{Z/p}$ action is all it takes to get a nontrivial covering (with base space a quotient of $L(p;q)$) – Max Jan 22 at 21:28
• This is a good advertisement for group actions on spaces. – Randall Jan 23 at 3:12
• @Randall : to be honest, I'm pretty new to the "algebraic topology game", so I don't know that many examples of interesting group actions on spaces - would you have more advertisement to give ? – Max Jan 27 at 22:02

I came up with an answer but I am not very sure if it is 100% correct. We will use two "heavy" theorems (while the post by Max uses none):

1. That the Galois Correspondance between coverings of $$X$$ and subgroups of $$π_1(Χ,x_0)$$ holds (since $$X$$ is a CW complex).
2. We know the fundamental groups of compact manifolds of dimension $$2$$ and more specifically, only $$\mathbb{R}P^2$$ has a finite fundametal group.

First, lets note that $$X$$ is compact since $$f$$ is onto and also a $$2$$-dim manifolds since $$f$$ is also a local homoemorphism.Now, let's call $$c:S^2\rightarrow \mathbb{R}P^2$$ the $$2$$-sheeted covering map. From what Max said in his/her answer, $$f\circ c$$ is a covering map and since $$c$$ is finite sheeted (again from Max's answer).Therefore $$f\circ c$$ is a finite sheeted covering from the $$S^2\rightarrow X$$. But from point $$2$$, if $$X$$ is not $$\mathbb{R}P^2$$ then $$π_1(X)$$ is infinite thereofore the trivial subgroup (which is the image of ($$π_1(S^2)$$ under any covering) has infinite sheets which leads to a contradiction. So $$X$$ is $$\mathbb{R}P^2$$.

But from the Galois Correspondance only $$S^2$$ and $$\mathbb{R}P^2$$ can cover $$\mathbb{R}P^2$$ and the latter does so via the identity. So $$c$$ is a homomorphism.

EDIT:I thought of another answer, even shorter but again uses theorem from covering space theory: $$\mathbb{R}P^2$$ has the fixed point property , meaning that every map $$g: \mathbb{R}P^2 \rightarrow \mathbb{R}P^2$$ has a fixed points (this can be proved with degree theory by lifting the maps between projective planes to maps between spheres). This means that the only deck transformation of your covering $$f$$ is the identity. Using Proposition 1.39 from Hatcer's book, we see that $$π_1(X)=\mathbb{Z}_2$$. Since the induced mapping $$f_*$$ is always injective for covering maps,we see that $$\forall x \in X : f^{-1}(x)=[π_1(X):f_*(π_1(\mathbb{R}P^2)]=1$$ (this is the index) therefore $$f$$ is a bijection and local homeomorphism which implies that $$f$$ is a homeomorphism.

• Good and short answer but the 2 written points or theorems was unknown for me but seems useful – pershina olad Feb 2 at 21:31
• Regarding your edit and your last paragraph, I don't see how the abelianness of $\pi_1(\mathbb{R}P^2)$ comes into play. If you have only one deck transformation, then $\pi_1(X)/p_*\pi_1(\mathbb{R}P^2)$ is of size $1$; and then you can conclude the same way. [Also, I don't know enough degree theory to understand your point; but that's on me] – Max Feb 6 at 9:04
• @Max If you have $A$ covers $B$ with $p$ and $D(A)$ the deck transformations, to guarantee that $D(A)=π_1(B)/p_*(π_1(Α))$ you need $A$ to be normal cover ,meaning $p_*(π_1(Α)$ is a normal subgroup. For the fixed point thing, see Exc 2.2.2 at page 155 of Hatcher's book. Hope this clarifies my edit :) – Nick A. Feb 6 at 9:17
• Right, my bad; but then it's not the abelianness of the covering space that matters, rather it's the one of the base space, isn't it ? And you can't predict when a covering will be normal, based only on the fundamental group of the covering space – Max Feb 6 at 9:43
• @Max Yes , I got confused but now it seems obvious that what i wrote is wrong. What I thought was something along the lines : If $A$ has the fixed point property and covers $B$ which has an abelian fundamental group then $A=B$. But this isn't so exciting to point out so I erased my last paragraph. – Nick A. Feb 6 at 10:00