Definition of 'simply connected'
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In the book 'Lie Groups, Lie Algebras, and Representations' written by Brian C. Hall, a matrix Lie group G is 'simply connected' if it is path-connected and for every continuous path $A(t),0le tle 1$, lying in G and with $A(0)=A(1)$, there exists a continuous function $A(s,t),0le s,tle 1$, taking values in G and having the following properties: (1) $A(s,0)=A(s,1)$ for all s, (2) $A(0,t)=A(t)$, and (3) $A(1,t)=A(1,0)$ for all t.
But on topology textbooks, it needs to choose a fixed base point, and the loop converges to that base point. While in the previous book, the loop can converge to any point in the loop and a base point is not needed. Are these two definitions equivalent?
algebraic-topology definition
edited Jan 15 at 6:33
José Carlos Santos
159k22126229
asked Jan 15 at 6:20
ZWJZWJ
33
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add a comment |
$begingroup$
In the book 'Lie Groups, Lie Algebras, and Representations' written by Brian C. Hall, a matrix Lie group G is 'simply connected' if it is path-connected and for every continuous path $A(t),0le tle 1$, lying in G and with $A(0)=A(1)$, there exists a continuous function $A(s,t),0le s,tle 1$, taking values in G and having the following properties: (1) $A(s,0)=A(s,1)$ for all s, (2) $A(0,t)=A(t)$, and (3) $A(1,t)=A(1,0)$ for all t.
But on topology textbooks, it needs to choose a fixed base point, and the loop converges to that base point. While in the previous book, the loop can converge to any point in the loop and a base point is not needed. Are these two definitions equivalent?
algebraic-topology definition
edited Jan 15 at 6:33
José Carlos Santos
159k22126229
asked Jan 15 at 6:20
ZWJZWJ
33
$endgroup$
$begingroup$
your space is path connected so the choice of base point isn't important since all the fundamental groups are isomorphic.
$endgroup$
– Bey Alexander
Jan 15 at 6:33
add a comment |
$begingroup$
In the book 'Lie Groups, Lie Algebras, and Representations' written by Brian C. Hall, a matrix Lie group G is 'simply connected' if it is path-connected and for every continuous path $A(t),0le tle 1$, lying in G and with $A(0)=A(1)$, there exists a continuous function $A(s,t),0le s,tle 1$, taking values in G and having the following properties: (1) $A(s,0)=A(s,1)$ for all s, (2) $A(0,t)=A(t)$, and (3) $A(1,t)=A(1,0)$ for all t.
But on topology textbooks, it needs to choose a fixed base point, and the loop converges to that base point. While in the previous book, the loop can converge to any point in the loop and a base point is not needed. Are these two definitions equivalent?
algebraic-topology definition
edited Jan 15 at 6:33
José Carlos Santos
159k22126229
asked Jan 15 at 6:20
ZWJZWJ
33
$endgroup$
In the book 'Lie Groups, Lie Algebras, and Representations' written by Brian C. Hall, a matrix Lie group G is 'simply connected' if it is path-connected and for every continuous path $A(t),0le tle 1$, lying in G and with $A(0)=A(1)$, there exists a continuous function $A(s,t),0le s,tle 1$, taking values in G and having the following properties: (1) $A(s,0)=A(s,1)$ for all s, (2) $A(0,t)=A(t)$, and (3) $A(1,t)=A(1,0)$ for all t.
But on topology textbooks, it needs to choose a fixed base point, and the loop converges to that base point. While in the previous book, the loop can converge to any point in the loop and a base point is not needed. Are these two definitions equivalent?
algebraic-topology definition
algebraic-topology definition
edited Jan 15 at 6:33
José Carlos Santos
159k22126229
asked Jan 15 at 6:20
ZWJZWJ
33
edited Jan 15 at 6:33
José Carlos Santos
159k22126229
asked Jan 15 at 6:20
ZWJZWJ
33
edited Jan 15 at 6:33
José Carlos Santos
159k22126229
edited Jan 15 at 6:33
José Carlos Santos
159k22126229
edited Jan 15 at 6:33
José Carlos Santos
159k22126229
159k22126229
asked Jan 15 at 6:20
ZWJZWJ
33
asked Jan 15 at 6:20
ZWJZWJ
33
asked Jan 15 at 6:20
ZWJZWJ
33
33
$begingroup$
your space is path connected so the choice of base point isn't important since all the fundamental groups are isomorphic.
$endgroup$
– Bey Alexander
Jan 15 at 6:33
add a comment |
$begingroup$
your space is path connected so the choice of base point isn't important since all the fundamental groups are isomorphic.
$endgroup$
– Bey Alexander
Jan 15 at 6:33
$begingroup$
your space is path connected so the choice of base point isn't important since all the fundamental groups are isomorphic.
$endgroup$
– Bey Alexander
Jan 15 at 6:33
$begingroup$
your space is path connected so the choice of base point isn't important since all the fundamental groups are isomorphic.
$endgroup$
– Bey Alexander
Jan 15 at 6:33
add a comment |
1 Answer
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Yes, they are equivalent under the assumption that the space is path-connected. Of course, in general they are not equivalent.
answered Jan 15 at 6:32
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
add a comment |
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Yes, they are equivalent under the assumption that the space is path-connected. Of course, in general they are not equivalent.
answered Jan 15 at 6:32
José Carlos SantosJosé Carlos Santos
159k22126229
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add a comment |
$begingroup$
Yes, they are equivalent under the assumption that the space is path-connected. Of course, in general they are not equivalent.
answered Jan 15 at 6:32
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
add a comment |
$begingroup$
Yes, they are equivalent under the assumption that the space is path-connected. Of course, in general they are not equivalent.
answered Jan 15 at 6:32
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
Yes, they are equivalent under the assumption that the space is path-connected. Of course, in general they are not equivalent.
answered Jan 15 at 6:32
José Carlos SantosJosé Carlos Santos
159k22126229
answered Jan 15 at 6:32
José Carlos SantosJosé Carlos Santos
159k22126229
answered Jan 15 at 6:32
José Carlos SantosJosé Carlos Santos
159k22126229
answered Jan 15 at 6:32
José Carlos SantosJosé Carlos Santos
159k22126229
159k22126229
add a comment |
add a comment |
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$begingroup$
your space is path connected so the choice of base point isn't important since all the fundamental groups are isomorphic.
$endgroup$
– Bey Alexander
Jan 15 at 6:33