This covering map is homeomorphism

Question

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

Answer 1

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.

Answer 2

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.