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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

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

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    $\begingroup$ 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? $\endgroup$ – Lukas Jan 22 at 19:47
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    $\begingroup$ @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) $\endgroup$ – Max Jan 22 at 21:25
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    $\begingroup$ 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 $L(p;q)$) $\endgroup$ – Max Jan 22 at 21:28
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    $\begingroup$ This is a good advertisement for group actions on spaces. $\endgroup$ – Randall Jan 23 at 3:12
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    $\begingroup$ @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 ? $\endgroup$ – Max Jan 27 at 22:02
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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.

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  • $\begingroup$ Good and short answer but the 2 written points or theorems was unknown for me but seems useful $\endgroup$ – pershina olad Feb 2 at 21:31
  • $\begingroup$ 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] $\endgroup$ – Max Feb 6 at 9:04
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    $\begingroup$ @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 :) $\endgroup$ – Nick A. Feb 6 at 9:17
  • $\begingroup$ 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 $\endgroup$ – Max Feb 6 at 9:43
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    $\begingroup$ @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. $\endgroup$ – Nick A. Feb 6 at 10:00

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