Mixing
Covering space with transitive action is path-connected.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I have this question:
Let $pcolon Erightarrow X$ with monodromy $lcolon pi_1(X)times p^-1(x_0)rightarrow p^-1(x_0)$ for a $x_0in X$.
Suppose that $X$ is path-connected. Show that $E$ is path-connected iff $l$ is transtive.
I think that if $E$ is path-connected, then for any $x,yin p^-1(x_0)$, there is a path $alpha$ such that $alpha(0)=x$ and $alpha(1)=y$, so considering the element $[p_ast(alpha)]in pi_1(X)$, this must do the action $p_ast(alpha)(x)=y$. Is this true?
For the other part, when $l$ is transitive, I think that if that's true, then for any $x,yin p^-1(x_0)$, there is an element $[alpha]in pi_1(X,x_0)$ such that $[alpha]x=y$, but I don't know why this element gives me a path from $x$ to $y$ in $E$. Could anyone explain this part to me and correct me if I'm wrong in any part?
proof-verification algebraic-topology covering-spaces fundamental-groups
asked Jul 17 at 10:22
MonsieurGalois
3,3961233
add a comment |Â
up vote
1
down vote
favorite
I have this question:
Let $pcolon Erightarrow X$ with monodromy $lcolon pi_1(X)times p^-1(x_0)rightarrow p^-1(x_0)$ for a $x_0in X$.
Suppose that $X$ is path-connected. Show that $E$ is path-connected iff $l$ is transtive.
I think that if $E$ is path-connected, then for any $x,yin p^-1(x_0)$, there is a path $alpha$ such that $alpha(0)=x$ and $alpha(1)=y$, so considering the element $[p_ast(alpha)]in pi_1(X)$, this must do the action $p_ast(alpha)(x)=y$. Is this true?
For the other part, when $l$ is transitive, I think that if that's true, then for any $x,yin p^-1(x_0)$, there is an element $[alpha]in pi_1(X,x_0)$ such that $[alpha]x=y$, but I don't know why this element gives me a path from $x$ to $y$ in $E$. Could anyone explain this part to me and correct me if I'm wrong in any part?
proof-verification algebraic-topology covering-spaces fundamental-groups
asked Jul 17 at 10:22
MonsieurGalois
3,3961233
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have this question:
Let $pcolon Erightarrow X$ with monodromy $lcolon pi_1(X)times p^-1(x_0)rightarrow p^-1(x_0)$ for a $x_0in X$.
Suppose that $X$ is path-connected. Show that $E$ is path-connected iff $l$ is transtive.
I think that if $E$ is path-connected, then for any $x,yin p^-1(x_0)$, there is a path $alpha$ such that $alpha(0)=x$ and $alpha(1)=y$, so considering the element $[p_ast(alpha)]in pi_1(X)$, this must do the action $p_ast(alpha)(x)=y$. Is this true?
For the other part, when $l$ is transitive, I think that if that's true, then for any $x,yin p^-1(x_0)$, there is an element $[alpha]in pi_1(X,x_0)$ such that $[alpha]x=y$, but I don't know why this element gives me a path from $x$ to $y$ in $E$. Could anyone explain this part to me and correct me if I'm wrong in any part?
proof-verification algebraic-topology covering-spaces fundamental-groups
asked Jul 17 at 10:22
MonsieurGalois
3,3961233
I have this question:
Let $pcolon Erightarrow X$ with monodromy $lcolon pi_1(X)times p^-1(x_0)rightarrow p^-1(x_0)$ for a $x_0in X$.
Suppose that $X$ is path-connected. Show that $E$ is path-connected iff $l$ is transtive.
I think that if $E$ is path-connected, then for any $x,yin p^-1(x_0)$, there is a path $alpha$ such that $alpha(0)=x$ and $alpha(1)=y$, so considering the element $[p_ast(alpha)]in pi_1(X)$, this must do the action $p_ast(alpha)(x)=y$. Is this true?
For the other part, when $l$ is transitive, I think that if that's true, then for any $x,yin p^-1(x_0)$, there is an element $[alpha]in pi_1(X,x_0)$ such that $[alpha]x=y$, but I don't know why this element gives me a path from $x$ to $y$ in $E$. Could anyone explain this part to me and correct me if I'm wrong in any part?
proof-verification algebraic-topology covering-spaces fundamental-groups
asked Jul 17 at 10:22
MonsieurGalois
3,3961233
asked Jul 17 at 10:22
MonsieurGalois
3,3961233
asked Jul 17 at 10:22
MonsieurGalois
3,3961233
asked Jul 17 at 10:22
MonsieurGalois
3,3961233
3,3961233
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
The first part is correct ($beta = p circ alpha$ is a loop in $(X,x_0)$ such that $[beta] cdot x = y$).
For the converse let $y_0 in p^-1(x_0)$. For any $y in E$ we shall construct a path from $y$ to $y_0$. Let $x = p(y) in X$. Choose a path in $X$ from $x$ to $x_0$. It has a lift to a path from $y$ to some $y_1 in p^-1(x_0)$. Next choose a loop $alpha$ in $(X,x_0)$ such that $[alpha] cdot y_1 = y_0$. This lifts to a path from $y_1$ to $y_0$.
answered Jul 17 at 13:29
Paul Frost
3,703420
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The first part is correct ($beta = p circ alpha$ is a loop in $(X,x_0)$ such that $[beta] cdot x = y$).
For the converse let $y_0 in p^-1(x_0)$. For any $y in E$ we shall construct a path from $y$ to $y_0$. Let $x = p(y) in X$. Choose a path in $X$ from $x$ to $x_0$. It has a lift to a path from $y$ to some $y_1 in p^-1(x_0)$. Next choose a loop $alpha$ in $(X,x_0)$ such that $[alpha] cdot y_1 = y_0$. This lifts to a path from $y_1$ to $y_0$.
answered Jul 17 at 13:29
Paul Frost
3,703420
add a comment |Â
up vote
1
down vote
The first part is correct ($beta = p circ alpha$ is a loop in $(X,x_0)$ such that $[beta] cdot x = y$).
For the converse let $y_0 in p^-1(x_0)$. For any $y in E$ we shall construct a path from $y$ to $y_0$. Let $x = p(y) in X$. Choose a path in $X$ from $x$ to $x_0$. It has a lift to a path from $y$ to some $y_1 in p^-1(x_0)$. Next choose a loop $alpha$ in $(X,x_0)$ such that $[alpha] cdot y_1 = y_0$. This lifts to a path from $y_1$ to $y_0$.
answered Jul 17 at 13:29
Paul Frost
3,703420
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The first part is correct ($beta = p circ alpha$ is a loop in $(X,x_0)$ such that $[beta] cdot x = y$).
For the converse let $y_0 in p^-1(x_0)$. For any $y in E$ we shall construct a path from $y$ to $y_0$. Let $x = p(y) in X$. Choose a path in $X$ from $x$ to $x_0$. It has a lift to a path from $y$ to some $y_1 in p^-1(x_0)$. Next choose a loop $alpha$ in $(X,x_0)$ such that $[alpha] cdot y_1 = y_0$. This lifts to a path from $y_1$ to $y_0$.
answered Jul 17 at 13:29
Paul Frost
3,703420
The first part is correct ($beta = p circ alpha$ is a loop in $(X,x_0)$ such that $[beta] cdot x = y$).
For the converse let $y_0 in p^-1(x_0)$. For any $y in E$ we shall construct a path from $y$ to $y_0$. Let $x = p(y) in X$. Choose a path in $X$ from $x$ to $x_0$. It has a lift to a path from $y$ to some $y_1 in p^-1(x_0)$. Next choose a loop $alpha$ in $(X,x_0)$ such that $[alpha] cdot y_1 = y_0$. This lifts to a path from $y_1$ to $y_0$.
answered Jul 17 at 13:29
Paul Frost
3,703420
edited Jul 17 at 20:13
answered Jul 17 at 13:29
Paul Frost
3,703420
answered Jul 17 at 13:29
Paul Frost
3,703420
answered Jul 17 at 13:29
Paul Frost
3,703420
3,703420
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2854366%2fcovering-space-with-transitive-action-is-path-connected%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password