Mixing
This covering map is homeomorphism
$begingroup$
Suppose that $f:mathbf RP^2rightarrow X$ 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
general-topology algebraic-topology complex-geometry covering-spaces
asked Jan 22 at 17:22
pershina oladpershina olad
9410
$endgroup$
|
show 1 more comment
$begingroup$
Suppose that $f:mathbf RP^2rightarrow X$ 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
general-topology algebraic-topology complex-geometry covering-spaces
asked Jan 22 at 17:22
pershina oladpershina olad
9410
$endgroup$
1
$begingroup$
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
1
$begingroup$
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
|
show 1 more comment
$begingroup$
Suppose that $f:mathbf RP^2rightarrow X$ 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
general-topology algebraic-topology complex-geometry covering-spaces
asked Jan 22 at 17:22
pershina oladpershina olad
9410
$endgroup$
Suppose that $f:mathbf RP^2rightarrow X$ 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
general-topology algebraic-topology complex-geometry covering-spaces
general-topology algebraic-topology complex-geometry covering-spaces
asked Jan 22 at 17:22
pershina oladpershina olad
9410
asked Jan 22 at 17:22
pershina oladpershina olad
9410
asked Jan 22 at 17:22
pershina oladpershina olad
9410
asked Jan 22 at 17:22
pershina oladpershina olad
9410
asked Jan 22 at 17:22
pershina oladpershina olad
9410
9410
1
$begingroup$
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
1
$begingroup$
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
|
show 1 more comment
1
$begingroup$
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
1
$begingroup$
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
1
1
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
@Max i think locally continuous gives us locally homeomorphism not homemorphism,am i right?
$endgroup$
– pershina olad
Jan 22 at 20:59
1
1
$begingroup$
Yes but a bijection which is a local homeomorphism is a homeomorphism
$endgroup$
– Max
Jan 22 at 21:15
$begingroup$
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
$begingroup$
o! I didnt know that.Thank you for your answer and good comments @Max
$endgroup$
– pershina olad
Jan 25 at 6:53
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
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 $gin G$, $d(g)$ 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 $gin G$ has no fixed points, then its action on $S^2$ is homotopic to $-id : S^2to S^2$ by $H(t,x) = frac{t gcdot x - (1-t)x}{||t gcdot x - (1-t)x||}$, this being well defined because by assumption, $t gcdot x - (1-t)x$ cannot vanish; so $g_* = (-id)_* = (-id)^3 = -id$, so $d(g) = -1$.
Thus if the action is free, and $gneq e$, then $d(g)neq 1$: $d$ is injective, thus providing the result.
2) If $p:Yto X$ 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 $q:Zto Y, p:Yto X$ are covering maps and $p$ is a finite sheeted covering, then $pcirc q$ 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 $p:Yto X$ 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 $pi : Yto Y/G$ 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 $xin X$, and the action will be the classical one, i.e. for some fixed $yin p^{-1}(x)$; for all $gin G$, 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 $xin X$. Then $p:Yto X$ is clearly $G$-invariant and so factors uniquely through some $f: Y/Gto X$. 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 $S^2to mathbf{R}P^2to X$ 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.
answered Jan 22 at 19:32
MaxMax
15.2k11143
$endgroup$
1
$begingroup$
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
1
$begingroup$
@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 $S^1to mathbf{R}P^1$ 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
2
$begingroup$
Say if you take a lens space $L(p;q)$ with $p,q$ coprime and odd, then $S^3$ should inherit a $mathbb{Z/ptimes Z/2}$ action of the form $(k,0)cdot (z_1,z_2) = (e^{frac{2ikpi}{p}}z_1 , e^{frac{2ikqpi}{p}}z_2)$ and $(k,1)cdot (z_1,z_2) = (-e^{frac{2ikpi}{p}}z_1 , -e^{frac{2ikqpi}{p}}z_2)$, 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
1
$begingroup$
This is a good advertisement for group actions on spaces.
$endgroup$
– Randall
Jan 23 at 3:12
1
$begingroup$
@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
add a comment |
$begingroup$
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):
- That the Galois Correspondance between coverings of $X$ and subgroups of $π_1(Χ,x_0)$ holds (since $X$ is a CW complex).
- 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 $c:S^2rightarrow mathbb{R}P^2$ the $2$-sheeted covering map. From what Max said in his/her answer, $fcirc c$ is a covering map and since $c$ is finite sheeted (again from Max's answer).Therefore $fcirc c$ is a finite sheeted covering from the $S^2rightarrow X$. 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.
answered Jan 30 at 14:17
Nick A.Nick A.
1,4311320
$endgroup$
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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
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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
1
$begingroup$
@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
1
$begingroup$
@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
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $gin G$, $d(g)$ 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 $gin G$ has no fixed points, then its action on $S^2$ is homotopic to $-id : S^2to S^2$ by $H(t,x) = frac{t gcdot x - (1-t)x}{||t gcdot x - (1-t)x||}$, this being well defined because by assumption, $t gcdot x - (1-t)x$ cannot vanish; so $g_* = (-id)_* = (-id)^3 = -id$, so $d(g) = -1$.
Thus if the action is free, and $gneq e$, then $d(g)neq 1$: $d$ is injective, thus providing the result.
2) If $p:Yto X$ 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 $q:Zto Y, p:Yto X$ are covering maps and $p$ is a finite sheeted covering, then $pcirc q$ 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 $p:Yto X$ 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 $pi : Yto Y/G$ 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 $xin X$, and the action will be the classical one, i.e. for some fixed $yin p^{-1}(x)$; for all $gin G$, 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 $xin X$. Then $p:Yto X$ is clearly $G$-invariant and so factors uniquely through some $f: Y/Gto X$. 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 $S^2to mathbf{R}P^2to X$ 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.
answered Jan 22 at 19:32
MaxMax
15.2k11143
$endgroup$
1
$begingroup$
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
1
$begingroup$
@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 $S^1to mathbf{R}P^1$ 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
2
$begingroup$
Say if you take a lens space $L(p;q)$ with $p,q$ coprime and odd, then $S^3$ should inherit a $mathbb{Z/ptimes Z/2}$ action of the form $(k,0)cdot (z_1,z_2) = (e^{frac{2ikpi}{p}}z_1 , e^{frac{2ikqpi}{p}}z_2)$ and $(k,1)cdot (z_1,z_2) = (-e^{frac{2ikpi}{p}}z_1 , -e^{frac{2ikqpi}{p}}z_2)$, 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
1
$begingroup$
This is a good advertisement for group actions on spaces.
$endgroup$
– Randall
Jan 23 at 3:12
1
$begingroup$
@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
add a comment |
$begingroup$
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 $gin G$, $d(g)$ 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 $gin G$ has no fixed points, then its action on $S^2$ is homotopic to $-id : S^2to S^2$ by $H(t,x) = frac{t gcdot x - (1-t)x}{||t gcdot x - (1-t)x||}$, this being well defined because by assumption, $t gcdot x - (1-t)x$ cannot vanish; so $g_* = (-id)_* = (-id)^3 = -id$, so $d(g) = -1$.
Thus if the action is free, and $gneq e$, then $d(g)neq 1$: $d$ is injective, thus providing the result.
2) If $p:Yto X$ 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 $q:Zto Y, p:Yto X$ are covering maps and $p$ is a finite sheeted covering, then $pcirc q$ 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 $p:Yto X$ 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 $pi : Yto Y/G$ 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 $xin X$, and the action will be the classical one, i.e. for some fixed $yin p^{-1}(x)$; for all $gin G$, 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 $xin X$. Then $p:Yto X$ is clearly $G$-invariant and so factors uniquely through some $f: Y/Gto X$. 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 $S^2to mathbf{R}P^2to X$ 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.
answered Jan 22 at 19:32
MaxMax
15.2k11143
$endgroup$
1
$begingroup$
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
1
$begingroup$
@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 $S^1to mathbf{R}P^1$ 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
2
$begingroup$
Say if you take a lens space $L(p;q)$ with $p,q$ coprime and odd, then $S^3$ should inherit a $mathbb{Z/ptimes Z/2}$ action of the form $(k,0)cdot (z_1,z_2) = (e^{frac{2ikpi}{p}}z_1 , e^{frac{2ikqpi}{p}}z_2)$ and $(k,1)cdot (z_1,z_2) = (-e^{frac{2ikpi}{p}}z_1 , -e^{frac{2ikqpi}{p}}z_2)$, 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
1
$begingroup$
This is a good advertisement for group actions on spaces.
$endgroup$
– Randall
Jan 23 at 3:12
1
$begingroup$
@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
add a comment |
$begingroup$
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 $gin G$, $d(g)$ 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 $gin G$ has no fixed points, then its action on $S^2$ is homotopic to $-id : S^2to S^2$ by $H(t,x) = frac{t gcdot x - (1-t)x}{||t gcdot x - (1-t)x||}$, this being well defined because by assumption, $t gcdot x - (1-t)x$ cannot vanish; so $g_* = (-id)_* = (-id)^3 = -id$, so $d(g) = -1$.
Thus if the action is free, and $gneq e$, then $d(g)neq 1$: $d$ is injective, thus providing the result.
2) If $p:Yto X$ 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 $q:Zto Y, p:Yto X$ are covering maps and $p$ is a finite sheeted covering, then $pcirc q$ 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 $p:Yto X$ 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 $pi : Yto Y/G$ 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 $xin X$, and the action will be the classical one, i.e. for some fixed $yin p^{-1}(x)$; for all $gin G$, 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 $xin X$. Then $p:Yto X$ is clearly $G$-invariant and so factors uniquely through some $f: Y/Gto X$. 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 $S^2to mathbf{R}P^2to X$ 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.
answered Jan 22 at 19:32
MaxMax
15.2k11143
$endgroup$
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 $gin G$, $d(g)$ 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 $gin G$ has no fixed points, then its action on $S^2$ is homotopic to $-id : S^2to S^2$ by $H(t,x) = frac{t gcdot x - (1-t)x}{||t gcdot x - (1-t)x||}$, this being well defined because by assumption, $t gcdot x - (1-t)x$ cannot vanish; so $g_* = (-id)_* = (-id)^3 = -id$, so $d(g) = -1$.
Thus if the action is free, and $gneq e$, then $d(g)neq 1$: $d$ is injective, thus providing the result.
2) If $p:Yto X$ 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 $q:Zto Y, p:Yto X$ are covering maps and $p$ is a finite sheeted covering, then $pcirc q$ 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 $p:Yto X$ 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 $pi : Yto Y/G$ 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 $xin X$, and the action will be the classical one, i.e. for some fixed $yin p^{-1}(x)$; for all $gin G$, 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 $xin X$. Then $p:Yto X$ is clearly $G$-invariant and so factors uniquely through some $f: Y/Gto X$. 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 $S^2to mathbf{R}P^2to X$ 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.
answered Jan 22 at 19:32
MaxMax
15.2k11143
answered Jan 22 at 19:32
MaxMax
15.2k11143
answered Jan 22 at 19:32
MaxMax
15.2k11143
answered Jan 22 at 19:32
MaxMax
15.2k11143
15.2k11143
1
$begingroup$
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
1
$begingroup$
@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 $S^1to mathbf{R}P^1$ 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
2
$begingroup$
Say if you take a lens space $L(p;q)$ with $p,q$ coprime and odd, then $S^3$ should inherit a $mathbb{Z/ptimes Z/2}$ action of the form $(k,0)cdot (z_1,z_2) = (e^{frac{2ikpi}{p}}z_1 , e^{frac{2ikqpi}{p}}z_2)$ and $(k,1)cdot (z_1,z_2) = (-e^{frac{2ikpi}{p}}z_1 , -e^{frac{2ikqpi}{p}}z_2)$, 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
1
$begingroup$
This is a good advertisement for group actions on spaces.
$endgroup$
– Randall
Jan 23 at 3:12
1
$begingroup$
@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
add a comment |
1
$begingroup$
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
1
$begingroup$
@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 $S^1to mathbf{R}P^1$ 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
2
$begingroup$
Say if you take a lens space $L(p;q)$ with $p,q$ coprime and odd, then $S^3$ should inherit a $mathbb{Z/ptimes Z/2}$ action of the form $(k,0)cdot (z_1,z_2) = (e^{frac{2ikpi}{p}}z_1 , e^{frac{2ikqpi}{p}}z_2)$ and $(k,1)cdot (z_1,z_2) = (-e^{frac{2ikpi}{p}}z_1 , -e^{frac{2ikqpi}{p}}z_2)$, 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
1
$begingroup$
This is a good advertisement for group actions on spaces.
$endgroup$
– Randall
Jan 23 at 3:12
1
$begingroup$
@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
1
1
$begingroup$
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
$begingroup$
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
1
1
$begingroup$
@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 $S^1to mathbf{R}P^1$ 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
$begingroup$
@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 $S^1to mathbf{R}P^1$ 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
2
2
$begingroup$
Say if you take a lens space $L(p;q)$ with $p,q$ coprime and odd, then $S^3$ should inherit a $mathbb{Z/ptimes Z/2}$ action of the form $(k,0)cdot (z_1,z_2) = (e^{frac{2ikpi}{p}}z_1 , e^{frac{2ikqpi}{p}}z_2)$ and $(k,1)cdot (z_1,z_2) = (-e^{frac{2ikpi}{p}}z_1 , -e^{frac{2ikqpi}{p}}z_2)$, 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
$begingroup$
Say if you take a lens space $L(p;q)$ with $p,q$ coprime and odd, then $S^3$ should inherit a $mathbb{Z/ptimes Z/2}$ action of the form $(k,0)cdot (z_1,z_2) = (e^{frac{2ikpi}{p}}z_1 , e^{frac{2ikqpi}{p}}z_2)$ and $(k,1)cdot (z_1,z_2) = (-e^{frac{2ikpi}{p}}z_1 , -e^{frac{2ikqpi}{p}}z_2)$, 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
1
1
$begingroup$
This is a good advertisement for group actions on spaces.
$endgroup$
– Randall
Jan 23 at 3:12
$begingroup$
This is a good advertisement for group actions on spaces.
$endgroup$
– Randall
Jan 23 at 3:12
1
1
$begingroup$
@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
$begingroup$
@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
add a comment |
$begingroup$
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):
- That the Galois Correspondance between coverings of $X$ and subgroups of $π_1(Χ,x_0)$ holds (since $X$ is a CW complex).
- 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 $c:S^2rightarrow mathbb{R}P^2$ the $2$-sheeted covering map. From what Max said in his/her answer, $fcirc c$ is a covering map and since $c$ is finite sheeted (again from Max's answer).Therefore $fcirc c$ is a finite sheeted covering from the $S^2rightarrow X$. 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.
answered Jan 30 at 14:17
Nick A.Nick A.
1,4311320
$endgroup$
$begingroup$
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
1
$begingroup$
@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 :)
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– 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
1
$begingroup$
@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
add a comment |
$begingroup$
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):
- That the Galois Correspondance between coverings of $X$ and subgroups of $π_1(Χ,x_0)$ holds (since $X$ is a CW complex).
- 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 $c:S^2rightarrow mathbb{R}P^2$ the $2$-sheeted covering map. From what Max said in his/her answer, $fcirc c$ is a covering map and since $c$ is finite sheeted (again from Max's answer).Therefore $fcirc c$ is a finite sheeted covering from the $S^2rightarrow X$. 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.
answered Jan 30 at 14:17
Nick A.Nick A.
1,4311320
$endgroup$
$begingroup$
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
1
$begingroup$
@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
1
$begingroup$
@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
add a comment |
$begingroup$
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):
- That the Galois Correspondance between coverings of $X$ and subgroups of $π_1(Χ,x_0)$ holds (since $X$ is a CW complex).
- 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 $c:S^2rightarrow mathbb{R}P^2$ the $2$-sheeted covering map. From what Max said in his/her answer, $fcirc c$ is a covering map and since $c$ is finite sheeted (again from Max's answer).Therefore $fcirc c$ is a finite sheeted covering from the $S^2rightarrow X$. 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.
answered Jan 30 at 14:17
Nick A.Nick A.
1,4311320
$endgroup$
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):
- That the Galois Correspondance between coverings of $X$ and subgroups of $π_1(Χ,x_0)$ holds (since $X$ is a CW complex).
- 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 $c:S^2rightarrow mathbb{R}P^2$ the $2$-sheeted covering map. From what Max said in his/her answer, $fcirc c$ is a covering map and since $c$ is finite sheeted (again from Max's answer).Therefore $fcirc c$ is a finite sheeted covering from the $S^2rightarrow X$. 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.
answered Jan 30 at 14:17
Nick A.Nick A.
1,4311320
edited Feb 6 at 9:55
answered Jan 30 at 14:17
Nick A.Nick A.
1,4311320
answered Jan 30 at 14:17
Nick A.Nick A.
1,4311320
answered Jan 30 at 14:17
Nick A.Nick A.
1,4311320
1,4311320
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Good and short answer but the 2 written points or theorems was unknown for me but seems useful
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– pershina olad
Feb 2 at 21:31
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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]
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– Max
Feb 6 at 9:04
1
$begingroup$
@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 :)
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– 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
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– Max
Feb 6 at 9:43
1
$begingroup$
@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.
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– Nick A.
Feb 6 at 10:00
add a comment |
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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
1
$begingroup$
@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
1
$begingroup$
@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
$begingroup$
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
$begingroup$
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
$begingroup$
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
1
1
$begingroup$
@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
$begingroup$
@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
$begingroup$
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
1
1
$begingroup$
@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
$begingroup$
@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
add a comment |
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All you need to get is 1-1ness, since $mathbb{R}P^2$ is compact and $X$ is Hausdorff.
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– Randall
Jan 22 at 17:49
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@Randall : or because the inverse (if it exists) is locally continuous, because $f$ is a covering map.
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– Max
Jan 22 at 18:26
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@Max i think locally continuous gives us locally homeomorphism not homemorphism,am i right?
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– pershina olad
Jan 22 at 20:59
1
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Yes but a bijection which is a local homeomorphism is a homeomorphism
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– Max
Jan 22 at 21:15
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o! I didnt know that.Thank you for your answer and good comments @Max
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– pershina olad
Jan 25 at 6:53