$pi_1(mathbbR^2setminus (mathbbZtimes 0))$
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I'm looking for the fundamental group of Y:= $mathbbR^2setminus (mathbbZtimes 0) $. In fact,I'm not. I already know that it is the free group in countable generators, but I'm looking for a proof.
My aim is to consider the rectangles $A_n=x-n$ because $Y=bigcup A_n $ and I want to use this corollary of Seifert van Kampen:
Let $X $ be a Hausdorff space, $A_icong mathbbS^1$ some subspaces of $ >X$ indexed by some set $ I$ such that
$X= undersetiin Ibigcup A_i $.
$A_icap A_j=x $ for every $i,jin I $ and for some $xin X $.
- A subset $Csubseteq X$ is open in $X$ (closed) if and only if $C cap A_i $ is open (closed) in $ A_i$ for every $iin I$.
Then the fundamental group of $X$ based on $x $ is the free group on $ I$
My idea is to retract each $A_n $ in to the circle centred in $n$ of radious $1/2 $ which is homeomorphic to the cirlce as required. Then I want to identify all the intersection points in the circles with a certain $x$ and take the quotient so that $A_n cap A_n =x$ for every $m,nin mathbbZ$. By doing this I think the conditions in the box are satisfied but I'm not sure if the resulting space is homeomorphic to $Y$ although it seems like it is.
If this is not the way I dont know which is. I'm really having troubles with this one. Please help.
Thank you all in advance
Important Note:
I'm aware of the fact that someone has already asked this questionbefore, but I'm not comfortably with the answer they gave him because it uses grupoids and other concepts which I think are not necessary since this is a question from the firsts chapters of a basic book in algebraic topology (namely Massey).
algebraic-topology fundamental-groups
asked Sep 3 at 14:03
Natalio
1359
add a comment |Â
up vote
1
down vote
favorite
I'm looking for the fundamental group of Y:= $mathbbR^2setminus (mathbbZtimes 0) $. In fact,I'm not. I already know that it is the free group in countable generators, but I'm looking for a proof.
My aim is to consider the rectangles $A_n=x-n$ because $Y=bigcup A_n $ and I want to use this corollary of Seifert van Kampen:
Let $X $ be a Hausdorff space, $A_icong mathbbS^1$ some subspaces of $ >X$ indexed by some set $ I$ such that
$X= undersetiin Ibigcup A_i $.
$A_icap A_j=x $ for every $i,jin I $ and for some $xin X $.
- A subset $Csubseteq X$ is open in $X$ (closed) if and only if $C cap A_i $ is open (closed) in $ A_i$ for every $iin I$.
Then the fundamental group of $X$ based on $x $ is the free group on $ I$
My idea is to retract each $A_n $ in to the circle centred in $n$ of radious $1/2 $ which is homeomorphic to the cirlce as required. Then I want to identify all the intersection points in the circles with a certain $x$ and take the quotient so that $A_n cap A_n =x$ for every $m,nin mathbbZ$. By doing this I think the conditions in the box are satisfied but I'm not sure if the resulting space is homeomorphic to $Y$ although it seems like it is.
If this is not the way I dont know which is. I'm really having troubles with this one. Please help.
Thank you all in advance
Important Note:
I'm aware of the fact that someone has already asked this questionbefore, but I'm not comfortably with the answer they gave him because it uses grupoids and other concepts which I think are not necessary since this is a question from the firsts chapters of a basic book in algebraic topology (namely Massey).
algebraic-topology fundamental-groups
asked Sep 3 at 14:03
Natalio
1359
I am not good with topology, but here is useful hint. Consider a real plane with a point missing. Mark fundamental group of that with $G$. Now divide plane with parallel lines $x + frac12$ for all $x in Z$. Obviously, each of those "stripes" is homeomorphic to first situation considered. Hence, result is $omega$ many copies of $G$
– nikola
Sep 3 at 14:32
maybe if you change your rectangles for something like $B_n = A_n cup y>1$, then you get that the intersections $B_i cap B_j$ are better behaved for the use of your theorem.
– jeanmfischer
Sep 3 at 14:51
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm looking for the fundamental group of Y:= $mathbbR^2setminus (mathbbZtimes 0) $. In fact,I'm not. I already know that it is the free group in countable generators, but I'm looking for a proof.
My aim is to consider the rectangles $A_n=x-n$ because $Y=bigcup A_n $ and I want to use this corollary of Seifert van Kampen:
Let $X $ be a Hausdorff space, $A_icong mathbbS^1$ some subspaces of $ >X$ indexed by some set $ I$ such that
$X= undersetiin Ibigcup A_i $.
$A_icap A_j=x $ for every $i,jin I $ and for some $xin X $.
- A subset $Csubseteq X$ is open in $X$ (closed) if and only if $C cap A_i $ is open (closed) in $ A_i$ for every $iin I$.
Then the fundamental group of $X$ based on $x $ is the free group on $ I$
My idea is to retract each $A_n $ in to the circle centred in $n$ of radious $1/2 $ which is homeomorphic to the cirlce as required. Then I want to identify all the intersection points in the circles with a certain $x$ and take the quotient so that $A_n cap A_n =x$ for every $m,nin mathbbZ$. By doing this I think the conditions in the box are satisfied but I'm not sure if the resulting space is homeomorphic to $Y$ although it seems like it is.
If this is not the way I dont know which is. I'm really having troubles with this one. Please help.
Thank you all in advance
Important Note:
I'm aware of the fact that someone has already asked this questionbefore, but I'm not comfortably with the answer they gave him because it uses grupoids and other concepts which I think are not necessary since this is a question from the firsts chapters of a basic book in algebraic topology (namely Massey).
algebraic-topology fundamental-groups
asked Sep 3 at 14:03
Natalio
1359
I'm looking for the fundamental group of Y:= $mathbbR^2setminus (mathbbZtimes 0) $. In fact,I'm not. I already know that it is the free group in countable generators, but I'm looking for a proof.
My aim is to consider the rectangles $A_n=x-n$ because $Y=bigcup A_n $ and I want to use this corollary of Seifert van Kampen:
Let $X $ be a Hausdorff space, $A_icong mathbbS^1$ some subspaces of $ >X$ indexed by some set $ I$ such that
$X= undersetiin Ibigcup A_i $.
$A_icap A_j=x $ for every $i,jin I $ and for some $xin X $.
- A subset $Csubseteq X$ is open in $X$ (closed) if and only if $C cap A_i $ is open (closed) in $ A_i$ for every $iin I$.
Then the fundamental group of $X$ based on $x $ is the free group on $ I$
My idea is to retract each $A_n $ in to the circle centred in $n$ of radious $1/2 $ which is homeomorphic to the cirlce as required. Then I want to identify all the intersection points in the circles with a certain $x$ and take the quotient so that $A_n cap A_n =x$ for every $m,nin mathbbZ$. By doing this I think the conditions in the box are satisfied but I'm not sure if the resulting space is homeomorphic to $Y$ although it seems like it is.
If this is not the way I dont know which is. I'm really having troubles with this one. Please help.
Thank you all in advance
Important Note:
I'm aware of the fact that someone has already asked this questionbefore, but I'm not comfortably with the answer they gave him because it uses grupoids and other concepts which I think are not necessary since this is a question from the firsts chapters of a basic book in algebraic topology (namely Massey).
algebraic-topology fundamental-groups
algebraic-topology fundamental-groups
asked Sep 3 at 14:03
Natalio
1359
asked Sep 3 at 14:03
Natalio
1359
edited Sep 3 at 14:23
asked Sep 3 at 14:03
Natalio
1359
asked Sep 3 at 14:03
Natalio
1359
asked Sep 3 at 14:03
Natalio
1359
1359
I am not good with topology, but here is useful hint. Consider a real plane with a point missing. Mark fundamental group of that with $G$. Now divide plane with parallel lines $x + frac12$ for all $x in Z$. Obviously, each of those "stripes" is homeomorphic to first situation considered. Hence, result is $omega$ many copies of $G$
– nikola
Sep 3 at 14:32
maybe if you change your rectangles for something like $B_n = A_n cup y>1$, then you get that the intersections $B_i cap B_j$ are better behaved for the use of your theorem.
– jeanmfischer
Sep 3 at 14:51
add a comment |Â
I am not good with topology, but here is useful hint. Consider a real plane with a point missing. Mark fundamental group of that with $G$. Now divide plane with parallel lines $x + frac12$ for all $x in Z$. Obviously, each of those "stripes" is homeomorphic to first situation considered. Hence, result is $omega$ many copies of $G$
– nikola
Sep 3 at 14:32
maybe if you change your rectangles for something like $B_n = A_n cup y>1$, then you get that the intersections $B_i cap B_j$ are better behaved for the use of your theorem.
– jeanmfischer
Sep 3 at 14:51
I am not good with topology, but here is useful hint. Consider a real plane with a point missing. Mark fundamental group of that with $G$. Now divide plane with parallel lines $x + frac12$ for all $x in Z$. Obviously, each of those "stripes" is homeomorphic to first situation considered. Hence, result is $omega$ many copies of $G$
– nikola
Sep 3 at 14:32
I am not good with topology, but here is useful hint. Consider a real plane with a point missing. Mark fundamental group of that with $G$. Now divide plane with parallel lines $x + frac12$ for all $x in Z$. Obviously, each of those "stripes" is homeomorphic to first situation considered. Hence, result is $omega$ many copies of $G$
– nikola
Sep 3 at 14:32
maybe if you change your rectangles for something like $B_n = A_n cup y>1$, then you get that the intersections $B_i cap B_j$ are better behaved for the use of your theorem.
– jeanmfischer
Sep 3 at 14:51
maybe if you change your rectangles for something like $B_n = A_n cup y>1$, then you get that the intersections $B_i cap B_j$ are better behaved for the use of your theorem.
– jeanmfischer
Sep 3 at 14:51
add a comment |Â
1 Answer
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I think your union of circles is a deformation retract of $Y$, so the spaces are homotopy equivalent and have the same homotopy groups.
A slightly better deformation retract of $Y$ may be taking circles of radius $1/4$ and one of their joint tangent lines. Then the set you need to quotient out becomes contractible.
answered Sep 3 at 16:02
Kusma
3,355218
Yes it is a deformation retract, but I don't know the group of the union of the circles. But I do know the group of the space resulting by taking the quotient of the equivalence relation generated by the set of the points of intersections of the circles. Si I would like to show that this quotient is homeomorphic to the union of the circles. I think it is, but how to prove it.
– Natalio
Sep 3 at 22:26
The infinite chain of circles that you get by contracting can be continuously deformed to an infinite wedge sum of circles (try and visualize this for three circles, then add more), and we know that fundamental groups are additive under taking wedge sums.
– Sofie Verbeek
Sep 4 at 17:32
The idea with the circles of radius 1/4 works very well but Ithink it is not a deformation retract of the original space.
– Natalio
Sep 9 at 0:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think your union of circles is a deformation retract of $Y$, so the spaces are homotopy equivalent and have the same homotopy groups.
A slightly better deformation retract of $Y$ may be taking circles of radius $1/4$ and one of their joint tangent lines. Then the set you need to quotient out becomes contractible.
answered Sep 3 at 16:02
Kusma
3,355218
Yes it is a deformation retract, but I don't know the group of the union of the circles. But I do know the group of the space resulting by taking the quotient of the equivalence relation generated by the set of the points of intersections of the circles. Si I would like to show that this quotient is homeomorphic to the union of the circles. I think it is, but how to prove it.
– Natalio
Sep 3 at 22:26
The infinite chain of circles that you get by contracting can be continuously deformed to an infinite wedge sum of circles (try and visualize this for three circles, then add more), and we know that fundamental groups are additive under taking wedge sums.
– Sofie Verbeek
Sep 4 at 17:32
The idea with the circles of radius 1/4 works very well but Ithink it is not a deformation retract of the original space.
– Natalio
Sep 9 at 0:03
add a comment |Â
up vote
0
down vote
I think your union of circles is a deformation retract of $Y$, so the spaces are homotopy equivalent and have the same homotopy groups.
A slightly better deformation retract of $Y$ may be taking circles of radius $1/4$ and one of their joint tangent lines. Then the set you need to quotient out becomes contractible.
answered Sep 3 at 16:02
Kusma
3,355218
Yes it is a deformation retract, but I don't know the group of the union of the circles. But I do know the group of the space resulting by taking the quotient of the equivalence relation generated by the set of the points of intersections of the circles. Si I would like to show that this quotient is homeomorphic to the union of the circles. I think it is, but how to prove it.
– Natalio
Sep 3 at 22:26
The infinite chain of circles that you get by contracting can be continuously deformed to an infinite wedge sum of circles (try and visualize this for three circles, then add more), and we know that fundamental groups are additive under taking wedge sums.
– Sofie Verbeek
Sep 4 at 17:32
The idea with the circles of radius 1/4 works very well but Ithink it is not a deformation retract of the original space.
– Natalio
Sep 9 at 0:03
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think your union of circles is a deformation retract of $Y$, so the spaces are homotopy equivalent and have the same homotopy groups.
A slightly better deformation retract of $Y$ may be taking circles of radius $1/4$ and one of their joint tangent lines. Then the set you need to quotient out becomes contractible.
answered Sep 3 at 16:02
Kusma
3,355218
I think your union of circles is a deformation retract of $Y$, so the spaces are homotopy equivalent and have the same homotopy groups.
A slightly better deformation retract of $Y$ may be taking circles of radius $1/4$ and one of their joint tangent lines. Then the set you need to quotient out becomes contractible.
answered Sep 3 at 16:02
Kusma
3,355218
edited Sep 4 at 6:15
answered Sep 3 at 16:02
Kusma
3,355218
answered Sep 3 at 16:02
Kusma
3,355218
answered Sep 3 at 16:02
Kusma
3,355218
3,355218
Yes it is a deformation retract, but I don't know the group of the union of the circles. But I do know the group of the space resulting by taking the quotient of the equivalence relation generated by the set of the points of intersections of the circles. Si I would like to show that this quotient is homeomorphic to the union of the circles. I think it is, but how to prove it.
– Natalio
Sep 3 at 22:26
The infinite chain of circles that you get by contracting can be continuously deformed to an infinite wedge sum of circles (try and visualize this for three circles, then add more), and we know that fundamental groups are additive under taking wedge sums.
– Sofie Verbeek
Sep 4 at 17:32
The idea with the circles of radius 1/4 works very well but Ithink it is not a deformation retract of the original space.
– Natalio
Sep 9 at 0:03
add a comment |Â
Yes it is a deformation retract, but I don't know the group of the union of the circles. But I do know the group of the space resulting by taking the quotient of the equivalence relation generated by the set of the points of intersections of the circles. Si I would like to show that this quotient is homeomorphic to the union of the circles. I think it is, but how to prove it.
– Natalio
Sep 3 at 22:26
The infinite chain of circles that you get by contracting can be continuously deformed to an infinite wedge sum of circles (try and visualize this for three circles, then add more), and we know that fundamental groups are additive under taking wedge sums.
– Sofie Verbeek
Sep 4 at 17:32
The idea with the circles of radius 1/4 works very well but Ithink it is not a deformation retract of the original space.
– Natalio
Sep 9 at 0:03
Yes it is a deformation retract, but I don't know the group of the union of the circles. But I do know the group of the space resulting by taking the quotient of the equivalence relation generated by the set of the points of intersections of the circles. Si I would like to show that this quotient is homeomorphic to the union of the circles. I think it is, but how to prove it.
– Natalio
Sep 3 at 22:26
Yes it is a deformation retract, but I don't know the group of the union of the circles. But I do know the group of the space resulting by taking the quotient of the equivalence relation generated by the set of the points of intersections of the circles. Si I would like to show that this quotient is homeomorphic to the union of the circles. I think it is, but how to prove it.
– Natalio
Sep 3 at 22:26
The infinite chain of circles that you get by contracting can be continuously deformed to an infinite wedge sum of circles (try and visualize this for three circles, then add more), and we know that fundamental groups are additive under taking wedge sums.
– Sofie Verbeek
Sep 4 at 17:32
The infinite chain of circles that you get by contracting can be continuously deformed to an infinite wedge sum of circles (try and visualize this for three circles, then add more), and we know that fundamental groups are additive under taking wedge sums.
– Sofie Verbeek
Sep 4 at 17:32
The idea with the circles of radius 1/4 works very well but Ithink it is not a deformation retract of the original space.
– Natalio
Sep 9 at 0:03
The idea with the circles of radius 1/4 works very well but Ithink it is not a deformation retract of the original space.
– Natalio
Sep 9 at 0:03
add a comment |Â
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I am not good with topology, but here is useful hint. Consider a real plane with a point missing. Mark fundamental group of that with $G$. Now divide plane with parallel lines $x + frac12$ for all $x in Z$. Obviously, each of those "stripes" is homeomorphic to first situation considered. Hence, result is $omega$ many copies of $G$
– nikola
Sep 3 at 14:32
maybe if you change your rectangles for something like $B_n = A_n cup y>1$, then you get that the intersections $B_i cap B_j$ are better behaved for the use of your theorem.
– jeanmfischer
Sep 3 at 14:51