covering spaces and equivalency of these three propositions [closed]
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This question is really important for me since the answer will give me the a way for solving similar proofs at algebraic topology lessons,so i need your help..
I need to prove these following statements are equivalent :
Suppose that $P:Ylongrightarrow X $ be a covering space. Prove that these three statements are equivalent :
A) $ {P_*}(pi_1(Y,{y_0}))lhd pi_1(X,P(y_0))$
B) For any two $y_0,y_1in P^{-1}(x_0)$ we have $ {P_*}(pi_1(Y,{y_0}))= {P_*}(pi_1(Y,{y_1}))$
C) For any loop $omega$ in $X$, all of its lifts $widetildeomega$ in $Y$ are loops or none of them is a loop.
I know I should try to show the equivalency by conjugacy classes but still I can not prove it, really thank anyone who help me with solving this problem
algebraic-topology normal-subgroups covering-spaces fundamental-groups
$endgroup$
closed as off-topic by José Carlos Santos, Eevee Trainer, Cesareo, Adrian Keister, amWhy Jan 9 at 15:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, Cesareo, Adrian Keister, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
This question is really important for me since the answer will give me the a way for solving similar proofs at algebraic topology lessons,so i need your help..
I need to prove these following statements are equivalent :
Suppose that $P:Ylongrightarrow X $ be a covering space. Prove that these three statements are equivalent :
A) $ {P_*}(pi_1(Y,{y_0}))lhd pi_1(X,P(y_0))$
B) For any two $y_0,y_1in P^{-1}(x_0)$ we have $ {P_*}(pi_1(Y,{y_0}))= {P_*}(pi_1(Y,{y_1}))$
C) For any loop $omega$ in $X$, all of its lifts $widetildeomega$ in $Y$ are loops or none of them is a loop.
I know I should try to show the equivalency by conjugacy classes but still I can not prove it, really thank anyone who help me with solving this problem
algebraic-topology normal-subgroups covering-spaces fundamental-groups
$endgroup$
closed as off-topic by José Carlos Santos, Eevee Trainer, Cesareo, Adrian Keister, amWhy Jan 9 at 15:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, Cesareo, Adrian Keister, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What have you tried ? Do you know how to prove any of the implications ? (I would advise A) => B) => C) => A) )
$endgroup$
– Max
Jan 8 at 22:07
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I try to work on it with conjugacy classes but i wasn't successful to reach the answers
$endgroup$
– pershina olad
Jan 8 at 22:11
add a comment |
$begingroup$
This question is really important for me since the answer will give me the a way for solving similar proofs at algebraic topology lessons,so i need your help..
I need to prove these following statements are equivalent :
Suppose that $P:Ylongrightarrow X $ be a covering space. Prove that these three statements are equivalent :
A) $ {P_*}(pi_1(Y,{y_0}))lhd pi_1(X,P(y_0))$
B) For any two $y_0,y_1in P^{-1}(x_0)$ we have $ {P_*}(pi_1(Y,{y_0}))= {P_*}(pi_1(Y,{y_1}))$
C) For any loop $omega$ in $X$, all of its lifts $widetildeomega$ in $Y$ are loops or none of them is a loop.
I know I should try to show the equivalency by conjugacy classes but still I can not prove it, really thank anyone who help me with solving this problem
algebraic-topology normal-subgroups covering-spaces fundamental-groups
$endgroup$
This question is really important for me since the answer will give me the a way for solving similar proofs at algebraic topology lessons,so i need your help..
I need to prove these following statements are equivalent :
Suppose that $P:Ylongrightarrow X $ be a covering space. Prove that these three statements are equivalent :
A) $ {P_*}(pi_1(Y,{y_0}))lhd pi_1(X,P(y_0))$
B) For any two $y_0,y_1in P^{-1}(x_0)$ we have $ {P_*}(pi_1(Y,{y_0}))= {P_*}(pi_1(Y,{y_1}))$
C) For any loop $omega$ in $X$, all of its lifts $widetildeomega$ in $Y$ are loops or none of them is a loop.
I know I should try to show the equivalency by conjugacy classes but still I can not prove it, really thank anyone who help me with solving this problem
algebraic-topology normal-subgroups covering-spaces fundamental-groups
algebraic-topology normal-subgroups covering-spaces fundamental-groups
edited Jan 9 at 18:36
pershina olad
asked Jan 8 at 16:08
pershina oladpershina olad
9410
9410
closed as off-topic by José Carlos Santos, Eevee Trainer, Cesareo, Adrian Keister, amWhy Jan 9 at 15:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, Cesareo, Adrian Keister, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Eevee Trainer, Cesareo, Adrian Keister, amWhy Jan 9 at 15:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, Cesareo, Adrian Keister, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What have you tried ? Do you know how to prove any of the implications ? (I would advise A) => B) => C) => A) )
$endgroup$
– Max
Jan 8 at 22:07
$begingroup$
I try to work on it with conjugacy classes but i wasn't successful to reach the answers
$endgroup$
– pershina olad
Jan 8 at 22:11
add a comment |
$begingroup$
What have you tried ? Do you know how to prove any of the implications ? (I would advise A) => B) => C) => A) )
$endgroup$
– Max
Jan 8 at 22:07
$begingroup$
I try to work on it with conjugacy classes but i wasn't successful to reach the answers
$endgroup$
– pershina olad
Jan 8 at 22:11
$begingroup$
What have you tried ? Do you know how to prove any of the implications ? (I would advise A) => B) => C) => A) )
$endgroup$
– Max
Jan 8 at 22:07
$begingroup$
What have you tried ? Do you know how to prove any of the implications ? (I would advise A) => B) => C) => A) )
$endgroup$
– Max
Jan 8 at 22:07
$begingroup$
I try to work on it with conjugacy classes but i wasn't successful to reach the answers
$endgroup$
– pershina olad
Jan 8 at 22:11
$begingroup$
I try to work on it with conjugacy classes but i wasn't successful to reach the answers
$endgroup$
– pershina olad
Jan 8 at 22:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let me give you indications, rather than a full solution :
A) => B) : note that if $alpha $ is a path from $y_0$ to $y_1$, then $p_*(pi_1(Y,y_0))$ and $p_*(pi_1(Y,y_1))$ are conjugate with $p_*(alpha)$ (which is a loop based at $x_0$).
B)=> C) : if $omega$ lifts to a loop $tilde{omega}$, then it is in $p_*(pi_1(Y,y_0))$ for some $y_0$; and so for all $y_0$. What does it mean ?
C) => A) : let $alpha$ be a loop based at $x_0$, $beta$ a loop based at $y_0$, and consider $alpha p_*(beta)alpha^{-1}$. If you lift $alpha$ to a path $gamma$ starting at $y_0$, then you can lift $p_*(beta)$ to a loop $delta$ based at $gamma (1)$. Then what is $p_*(gammadeltagamma^{-1})$ ?
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Awesome answer thank you max,but in C==>A I'm not sure i understand ...what you asked for is a member at fundamental group of X at P(y0)? thank for your help and support Max . @Max
$endgroup$
– pershina olad
Jan 10 at 12:29
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$x_0 = p(y_0)$; what part are you talking about ?
$endgroup$
– Max
Jan 10 at 16:17
$begingroup$
I mean the last question you asked from me at the proof C==>A, p∗(γδγ−1) , the answer of your question is that it belongs to the fundamental group of X at P(y0)? @Max
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– pershina olad
Jan 11 at 6:45
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Yes but most importantly, what is its value ?
$endgroup$
– Max
Jan 11 at 10:30
$begingroup$
mmm I am a little weak in algebraic constructions because the field i study is dynamical system in geometry but i think the abelian property which gives the normal subgroup is your goal to reach me to the main concept,am i right? @Max
$endgroup$
– pershina olad
Jan 12 at 5:51
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let me give you indications, rather than a full solution :
A) => B) : note that if $alpha $ is a path from $y_0$ to $y_1$, then $p_*(pi_1(Y,y_0))$ and $p_*(pi_1(Y,y_1))$ are conjugate with $p_*(alpha)$ (which is a loop based at $x_0$).
B)=> C) : if $omega$ lifts to a loop $tilde{omega}$, then it is in $p_*(pi_1(Y,y_0))$ for some $y_0$; and so for all $y_0$. What does it mean ?
C) => A) : let $alpha$ be a loop based at $x_0$, $beta$ a loop based at $y_0$, and consider $alpha p_*(beta)alpha^{-1}$. If you lift $alpha$ to a path $gamma$ starting at $y_0$, then you can lift $p_*(beta)$ to a loop $delta$ based at $gamma (1)$. Then what is $p_*(gammadeltagamma^{-1})$ ?
$endgroup$
$begingroup$
Awesome answer thank you max,but in C==>A I'm not sure i understand ...what you asked for is a member at fundamental group of X at P(y0)? thank for your help and support Max . @Max
$endgroup$
– pershina olad
Jan 10 at 12:29
$begingroup$
$x_0 = p(y_0)$; what part are you talking about ?
$endgroup$
– Max
Jan 10 at 16:17
$begingroup$
I mean the last question you asked from me at the proof C==>A, p∗(γδγ−1) , the answer of your question is that it belongs to the fundamental group of X at P(y0)? @Max
$endgroup$
– pershina olad
Jan 11 at 6:45
$begingroup$
Yes but most importantly, what is its value ?
$endgroup$
– Max
Jan 11 at 10:30
$begingroup$
mmm I am a little weak in algebraic constructions because the field i study is dynamical system in geometry but i think the abelian property which gives the normal subgroup is your goal to reach me to the main concept,am i right? @Max
$endgroup$
– pershina olad
Jan 12 at 5:51
|
show 2 more comments
$begingroup$
Let me give you indications, rather than a full solution :
A) => B) : note that if $alpha $ is a path from $y_0$ to $y_1$, then $p_*(pi_1(Y,y_0))$ and $p_*(pi_1(Y,y_1))$ are conjugate with $p_*(alpha)$ (which is a loop based at $x_0$).
B)=> C) : if $omega$ lifts to a loop $tilde{omega}$, then it is in $p_*(pi_1(Y,y_0))$ for some $y_0$; and so for all $y_0$. What does it mean ?
C) => A) : let $alpha$ be a loop based at $x_0$, $beta$ a loop based at $y_0$, and consider $alpha p_*(beta)alpha^{-1}$. If you lift $alpha$ to a path $gamma$ starting at $y_0$, then you can lift $p_*(beta)$ to a loop $delta$ based at $gamma (1)$. Then what is $p_*(gammadeltagamma^{-1})$ ?
$endgroup$
$begingroup$
Awesome answer thank you max,but in C==>A I'm not sure i understand ...what you asked for is a member at fundamental group of X at P(y0)? thank for your help and support Max . @Max
$endgroup$
– pershina olad
Jan 10 at 12:29
$begingroup$
$x_0 = p(y_0)$; what part are you talking about ?
$endgroup$
– Max
Jan 10 at 16:17
$begingroup$
I mean the last question you asked from me at the proof C==>A, p∗(γδγ−1) , the answer of your question is that it belongs to the fundamental group of X at P(y0)? @Max
$endgroup$
– pershina olad
Jan 11 at 6:45
$begingroup$
Yes but most importantly, what is its value ?
$endgroup$
– Max
Jan 11 at 10:30
$begingroup$
mmm I am a little weak in algebraic constructions because the field i study is dynamical system in geometry but i think the abelian property which gives the normal subgroup is your goal to reach me to the main concept,am i right? @Max
$endgroup$
– pershina olad
Jan 12 at 5:51
|
show 2 more comments
$begingroup$
Let me give you indications, rather than a full solution :
A) => B) : note that if $alpha $ is a path from $y_0$ to $y_1$, then $p_*(pi_1(Y,y_0))$ and $p_*(pi_1(Y,y_1))$ are conjugate with $p_*(alpha)$ (which is a loop based at $x_0$).
B)=> C) : if $omega$ lifts to a loop $tilde{omega}$, then it is in $p_*(pi_1(Y,y_0))$ for some $y_0$; and so for all $y_0$. What does it mean ?
C) => A) : let $alpha$ be a loop based at $x_0$, $beta$ a loop based at $y_0$, and consider $alpha p_*(beta)alpha^{-1}$. If you lift $alpha$ to a path $gamma$ starting at $y_0$, then you can lift $p_*(beta)$ to a loop $delta$ based at $gamma (1)$. Then what is $p_*(gammadeltagamma^{-1})$ ?
$endgroup$
Let me give you indications, rather than a full solution :
A) => B) : note that if $alpha $ is a path from $y_0$ to $y_1$, then $p_*(pi_1(Y,y_0))$ and $p_*(pi_1(Y,y_1))$ are conjugate with $p_*(alpha)$ (which is a loop based at $x_0$).
B)=> C) : if $omega$ lifts to a loop $tilde{omega}$, then it is in $p_*(pi_1(Y,y_0))$ for some $y_0$; and so for all $y_0$. What does it mean ?
C) => A) : let $alpha$ be a loop based at $x_0$, $beta$ a loop based at $y_0$, and consider $alpha p_*(beta)alpha^{-1}$. If you lift $alpha$ to a path $gamma$ starting at $y_0$, then you can lift $p_*(beta)$ to a loop $delta$ based at $gamma (1)$. Then what is $p_*(gammadeltagamma^{-1})$ ?
answered Jan 8 at 23:04
MaxMax
14.7k11143
14.7k11143
$begingroup$
Awesome answer thank you max,but in C==>A I'm not sure i understand ...what you asked for is a member at fundamental group of X at P(y0)? thank for your help and support Max . @Max
$endgroup$
– pershina olad
Jan 10 at 12:29
$begingroup$
$x_0 = p(y_0)$; what part are you talking about ?
$endgroup$
– Max
Jan 10 at 16:17
$begingroup$
I mean the last question you asked from me at the proof C==>A, p∗(γδγ−1) , the answer of your question is that it belongs to the fundamental group of X at P(y0)? @Max
$endgroup$
– pershina olad
Jan 11 at 6:45
$begingroup$
Yes but most importantly, what is its value ?
$endgroup$
– Max
Jan 11 at 10:30
$begingroup$
mmm I am a little weak in algebraic constructions because the field i study is dynamical system in geometry but i think the abelian property which gives the normal subgroup is your goal to reach me to the main concept,am i right? @Max
$endgroup$
– pershina olad
Jan 12 at 5:51
|
show 2 more comments
$begingroup$
Awesome answer thank you max,but in C==>A I'm not sure i understand ...what you asked for is a member at fundamental group of X at P(y0)? thank for your help and support Max . @Max
$endgroup$
– pershina olad
Jan 10 at 12:29
$begingroup$
$x_0 = p(y_0)$; what part are you talking about ?
$endgroup$
– Max
Jan 10 at 16:17
$begingroup$
I mean the last question you asked from me at the proof C==>A, p∗(γδγ−1) , the answer of your question is that it belongs to the fundamental group of X at P(y0)? @Max
$endgroup$
– pershina olad
Jan 11 at 6:45
$begingroup$
Yes but most importantly, what is its value ?
$endgroup$
– Max
Jan 11 at 10:30
$begingroup$
mmm I am a little weak in algebraic constructions because the field i study is dynamical system in geometry but i think the abelian property which gives the normal subgroup is your goal to reach me to the main concept,am i right? @Max
$endgroup$
– pershina olad
Jan 12 at 5:51
$begingroup$
Awesome answer thank you max,but in C==>A I'm not sure i understand ...what you asked for is a member at fundamental group of X at P(y0)? thank for your help and support Max . @Max
$endgroup$
– pershina olad
Jan 10 at 12:29
$begingroup$
Awesome answer thank you max,but in C==>A I'm not sure i understand ...what you asked for is a member at fundamental group of X at P(y0)? thank for your help and support Max . @Max
$endgroup$
– pershina olad
Jan 10 at 12:29
$begingroup$
$x_0 = p(y_0)$; what part are you talking about ?
$endgroup$
– Max
Jan 10 at 16:17
$begingroup$
$x_0 = p(y_0)$; what part are you talking about ?
$endgroup$
– Max
Jan 10 at 16:17
$begingroup$
I mean the last question you asked from me at the proof C==>A, p∗(γδγ−1) , the answer of your question is that it belongs to the fundamental group of X at P(y0)? @Max
$endgroup$
– pershina olad
Jan 11 at 6:45
$begingroup$
I mean the last question you asked from me at the proof C==>A, p∗(γδγ−1) , the answer of your question is that it belongs to the fundamental group of X at P(y0)? @Max
$endgroup$
– pershina olad
Jan 11 at 6:45
$begingroup$
Yes but most importantly, what is its value ?
$endgroup$
– Max
Jan 11 at 10:30
$begingroup$
Yes but most importantly, what is its value ?
$endgroup$
– Max
Jan 11 at 10:30
$begingroup$
mmm I am a little weak in algebraic constructions because the field i study is dynamical system in geometry but i think the abelian property which gives the normal subgroup is your goal to reach me to the main concept,am i right? @Max
$endgroup$
– pershina olad
Jan 12 at 5:51
$begingroup$
mmm I am a little weak in algebraic constructions because the field i study is dynamical system in geometry but i think the abelian property which gives the normal subgroup is your goal to reach me to the main concept,am i right? @Max
$endgroup$
– pershina olad
Jan 12 at 5:51
|
show 2 more comments
$begingroup$
What have you tried ? Do you know how to prove any of the implications ? (I would advise A) => B) => C) => A) )
$endgroup$
– Max
Jan 8 at 22:07
$begingroup$
I try to work on it with conjugacy classes but i wasn't successful to reach the answers
$endgroup$
– pershina olad
Jan 8 at 22:11