Finite graph homotopically equivalent to wedge sum of finite circle.
$begingroup$
Show that if X is a finite graph i.e. a graph with finitely many vertices and finitely many
edges, then X is homotopically equivalent to wedge sum of finitely circles.
I know that I will be able to make the connected components of finite graph homotopically equivalents to wedge sum of finite circles as I can shrink few edges but how do I proceed further
algebraic-topology
asked Oct 16 '16 at 12:16
scotscot
62
$endgroup$
add a comment |
$begingroup$
Show that if X is a finite graph i.e. a graph with finitely many vertices and finitely many
edges, then X is homotopically equivalent to wedge sum of finitely circles.
I know that I will be able to make the connected components of finite graph homotopically equivalents to wedge sum of finite circles as I can shrink few edges but how do I proceed further
algebraic-topology
asked Oct 16 '16 at 12:16
scotscot
62
$endgroup$
$begingroup$
You have to show that contracting an edge $Gto G/e$ is a homotopy equivalence. This can be done by hand, and is surprisingly tricky. You need to define maps in both directions (in the definition of homotopy equivalence), and defining the map from $G/eto G$ is the tricky bit but if you think about it, there aren't that many ways to do it. You could also appeal to more general theorems that state that contracting contractible subspaces yield homotopy equivalences under favorable circumstances. It depends what theorems you have access to.
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– Cheerful Parsnip
Dec 1 '18 at 5:24
$begingroup$
See math.stackexchange.com/questions/21705/…
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 7:26
add a comment |
$begingroup$
Show that if X is a finite graph i.e. a graph with finitely many vertices and finitely many
edges, then X is homotopically equivalent to wedge sum of finitely circles.
I know that I will be able to make the connected components of finite graph homotopically equivalents to wedge sum of finite circles as I can shrink few edges but how do I proceed further
algebraic-topology
asked Oct 16 '16 at 12:16
scotscot
62
$endgroup$
Show that if X is a finite graph i.e. a graph with finitely many vertices and finitely many
edges, then X is homotopically equivalent to wedge sum of finitely circles.
I know that I will be able to make the connected components of finite graph homotopically equivalents to wedge sum of finite circles as I can shrink few edges but how do I proceed further
algebraic-topology
algebraic-topology
asked Oct 16 '16 at 12:16
scotscot
62
asked Oct 16 '16 at 12:16
scotscot
62
asked Oct 16 '16 at 12:16
scotscot
62
asked Oct 16 '16 at 12:16
scotscot
62
asked Oct 16 '16 at 12:16
scotscot
62
62
$begingroup$
You have to show that contracting an edge $Gto G/e$ is a homotopy equivalence. This can be done by hand, and is surprisingly tricky. You need to define maps in both directions (in the definition of homotopy equivalence), and defining the map from $G/eto G$ is the tricky bit but if you think about it, there aren't that many ways to do it. You could also appeal to more general theorems that state that contracting contractible subspaces yield homotopy equivalences under favorable circumstances. It depends what theorems you have access to.
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 5:24
$begingroup$
See math.stackexchange.com/questions/21705/…
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 7:26
add a comment |
$begingroup$
You have to show that contracting an edge $Gto G/e$ is a homotopy equivalence. This can be done by hand, and is surprisingly tricky. You need to define maps in both directions (in the definition of homotopy equivalence), and defining the map from $G/eto G$ is the tricky bit but if you think about it, there aren't that many ways to do it. You could also appeal to more general theorems that state that contracting contractible subspaces yield homotopy equivalences under favorable circumstances. It depends what theorems you have access to.
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 5:24
$begingroup$
See math.stackexchange.com/questions/21705/…
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 7:26
$begingroup$
You have to show that contracting an edge $Gto G/e$ is a homotopy equivalence. This can be done by hand, and is surprisingly tricky. You need to define maps in both directions (in the definition of homotopy equivalence), and defining the map from $G/eto G$ is the tricky bit but if you think about it, there aren't that many ways to do it. You could also appeal to more general theorems that state that contracting contractible subspaces yield homotopy equivalences under favorable circumstances. It depends what theorems you have access to.
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 5:24
$begingroup$
You have to show that contracting an edge $Gto G/e$ is a homotopy equivalence. This can be done by hand, and is surprisingly tricky. You need to define maps in both directions (in the definition of homotopy equivalence), and defining the map from $G/eto G$ is the tricky bit but if you think about it, there aren't that many ways to do it. You could also appeal to more general theorems that state that contracting contractible subspaces yield homotopy equivalences under favorable circumstances. It depends what theorems you have access to.
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 5:24
$begingroup$
See math.stackexchange.com/questions/21705/…
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 7:26
$begingroup$
See math.stackexchange.com/questions/21705/…
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 7:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Intuition goes something like this:
Let $X$ be a finite connected graph. Every tree is contractible, (deformation retractable to a point), therefore if $X$ is a tree then $X simeq 0$.
Otherwise, consider chordless cycles of $X$, (cycles that don't have subcycles). Any chordless cycle is homotopy equivalent to $S^1$.
Edge is a tree, so for any two chordless cycles sharing an edge, the edge contracts to a point, resulting in homotopy equivalency to $S^1 vee S^1$. Same goes for $n$ cycles. Therefore, everything besides chordless cycles contracts to a point $x_0$ at which all of $X$'s chordless cycles are wedged at.
Concatenation of trees and chordless cycles makes up any finite graph.
Therefore, any finite graph $X$ with $n$ chordless cycles:
$$X simeq underbrace{S^1 vee S^1 vee cdots vee S^1}_{ntext{ times}}$$
answered Feb 28 '18 at 22:47
Eetu KoskelaEetu Koskela
708
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add a comment |
$begingroup$
Similar idea as in Eetu Koskela's answer.
Just want to point out that one can refer to Example 1.22 and section 1.B in Hatcher's Algebraic Topology.
First look for a spanning tree of the connected graph. Then use Seifert-van Kampen theorem (where we deformation retract things), we can see that the fundamental group of the graph is a free group, say $F_n$. Take the wedge sum of $n$ circles, and its fundamental group is also $F_n$. Since both our graph and the wedge sum of finite circles (essentially also a connected graph, with loops) are CW complex $K(G,1)$ that have the same fundamental group, they are homotopy equivalent.
answered Dec 1 '18 at 5:09
chikurinchikurin
899
$endgroup$
$begingroup$
This is a bit high-powered given the likely context of what OP knows.
$endgroup$
– Joshua Mundinger
Dec 1 '18 at 5:26
$begingroup$
Probably yes... I was just searching for similar questions and kind of want to write an answer for myself or if by any chance it can help the community.
$endgroup$
– chikurin
Dec 1 '18 at 5:33
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Intuition goes something like this:
Let $X$ be a finite connected graph. Every tree is contractible, (deformation retractable to a point), therefore if $X$ is a tree then $X simeq 0$.
Otherwise, consider chordless cycles of $X$, (cycles that don't have subcycles). Any chordless cycle is homotopy equivalent to $S^1$.
Edge is a tree, so for any two chordless cycles sharing an edge, the edge contracts to a point, resulting in homotopy equivalency to $S^1 vee S^1$. Same goes for $n$ cycles. Therefore, everything besides chordless cycles contracts to a point $x_0$ at which all of $X$'s chordless cycles are wedged at.
Concatenation of trees and chordless cycles makes up any finite graph.
Therefore, any finite graph $X$ with $n$ chordless cycles:
$$X simeq underbrace{S^1 vee S^1 vee cdots vee S^1}_{ntext{ times}}$$
answered Feb 28 '18 at 22:47
Eetu KoskelaEetu Koskela
708
$endgroup$
add a comment |
$begingroup$
Intuition goes something like this:
Let $X$ be a finite connected graph. Every tree is contractible, (deformation retractable to a point), therefore if $X$ is a tree then $X simeq 0$.
Otherwise, consider chordless cycles of $X$, (cycles that don't have subcycles). Any chordless cycle is homotopy equivalent to $S^1$.
Edge is a tree, so for any two chordless cycles sharing an edge, the edge contracts to a point, resulting in homotopy equivalency to $S^1 vee S^1$. Same goes for $n$ cycles. Therefore, everything besides chordless cycles contracts to a point $x_0$ at which all of $X$'s chordless cycles are wedged at.
Concatenation of trees and chordless cycles makes up any finite graph.
Therefore, any finite graph $X$ with $n$ chordless cycles:
$$X simeq underbrace{S^1 vee S^1 vee cdots vee S^1}_{ntext{ times}}$$
answered Feb 28 '18 at 22:47
Eetu KoskelaEetu Koskela
708
$endgroup$
add a comment |
$begingroup$
Intuition goes something like this:
Let $X$ be a finite connected graph. Every tree is contractible, (deformation retractable to a point), therefore if $X$ is a tree then $X simeq 0$.
Otherwise, consider chordless cycles of $X$, (cycles that don't have subcycles). Any chordless cycle is homotopy equivalent to $S^1$.
Edge is a tree, so for any two chordless cycles sharing an edge, the edge contracts to a point, resulting in homotopy equivalency to $S^1 vee S^1$. Same goes for $n$ cycles. Therefore, everything besides chordless cycles contracts to a point $x_0$ at which all of $X$'s chordless cycles are wedged at.
Concatenation of trees and chordless cycles makes up any finite graph.
Therefore, any finite graph $X$ with $n$ chordless cycles:
$$X simeq underbrace{S^1 vee S^1 vee cdots vee S^1}_{ntext{ times}}$$
answered Feb 28 '18 at 22:47
Eetu KoskelaEetu Koskela
708
$endgroup$
Intuition goes something like this:
Let $X$ be a finite connected graph. Every tree is contractible, (deformation retractable to a point), therefore if $X$ is a tree then $X simeq 0$.
Otherwise, consider chordless cycles of $X$, (cycles that don't have subcycles). Any chordless cycle is homotopy equivalent to $S^1$.
Edge is a tree, so for any two chordless cycles sharing an edge, the edge contracts to a point, resulting in homotopy equivalency to $S^1 vee S^1$. Same goes for $n$ cycles. Therefore, everything besides chordless cycles contracts to a point $x_0$ at which all of $X$'s chordless cycles are wedged at.
Concatenation of trees and chordless cycles makes up any finite graph.
Therefore, any finite graph $X$ with $n$ chordless cycles:
$$X simeq underbrace{S^1 vee S^1 vee cdots vee S^1}_{ntext{ times}}$$
answered Feb 28 '18 at 22:47
Eetu KoskelaEetu Koskela
708
edited Feb 28 '18 at 23:25
answered Feb 28 '18 at 22:47
Eetu KoskelaEetu Koskela
708
answered Feb 28 '18 at 22:47
Eetu KoskelaEetu Koskela
708
answered Feb 28 '18 at 22:47
Eetu KoskelaEetu Koskela
708
708
add a comment |
add a comment |
$begingroup$
Similar idea as in Eetu Koskela's answer.
Just want to point out that one can refer to Example 1.22 and section 1.B in Hatcher's Algebraic Topology.
First look for a spanning tree of the connected graph. Then use Seifert-van Kampen theorem (where we deformation retract things), we can see that the fundamental group of the graph is a free group, say $F_n$. Take the wedge sum of $n$ circles, and its fundamental group is also $F_n$. Since both our graph and the wedge sum of finite circles (essentially also a connected graph, with loops) are CW complex $K(G,1)$ that have the same fundamental group, they are homotopy equivalent.
answered Dec 1 '18 at 5:09
chikurinchikurin
899
$endgroup$
$begingroup$
This is a bit high-powered given the likely context of what OP knows.
$endgroup$
– Joshua Mundinger
Dec 1 '18 at 5:26
$begingroup$
Probably yes... I was just searching for similar questions and kind of want to write an answer for myself or if by any chance it can help the community.
$endgroup$
– chikurin
Dec 1 '18 at 5:33
add a comment |
$begingroup$
Similar idea as in Eetu Koskela's answer.
Just want to point out that one can refer to Example 1.22 and section 1.B in Hatcher's Algebraic Topology.
First look for a spanning tree of the connected graph. Then use Seifert-van Kampen theorem (where we deformation retract things), we can see that the fundamental group of the graph is a free group, say $F_n$. Take the wedge sum of $n$ circles, and its fundamental group is also $F_n$. Since both our graph and the wedge sum of finite circles (essentially also a connected graph, with loops) are CW complex $K(G,1)$ that have the same fundamental group, they are homotopy equivalent.
answered Dec 1 '18 at 5:09
chikurinchikurin
899
$endgroup$
$begingroup$
This is a bit high-powered given the likely context of what OP knows.
$endgroup$
– Joshua Mundinger
Dec 1 '18 at 5:26
$begingroup$
Probably yes... I was just searching for similar questions and kind of want to write an answer for myself or if by any chance it can help the community.
$endgroup$
– chikurin
Dec 1 '18 at 5:33
add a comment |
$begingroup$
Similar idea as in Eetu Koskela's answer.
Just want to point out that one can refer to Example 1.22 and section 1.B in Hatcher's Algebraic Topology.
First look for a spanning tree of the connected graph. Then use Seifert-van Kampen theorem (where we deformation retract things), we can see that the fundamental group of the graph is a free group, say $F_n$. Take the wedge sum of $n$ circles, and its fundamental group is also $F_n$. Since both our graph and the wedge sum of finite circles (essentially also a connected graph, with loops) are CW complex $K(G,1)$ that have the same fundamental group, they are homotopy equivalent.
answered Dec 1 '18 at 5:09
chikurinchikurin
899
$endgroup$
Similar idea as in Eetu Koskela's answer.
Just want to point out that one can refer to Example 1.22 and section 1.B in Hatcher's Algebraic Topology.
First look for a spanning tree of the connected graph. Then use Seifert-van Kampen theorem (where we deformation retract things), we can see that the fundamental group of the graph is a free group, say $F_n$. Take the wedge sum of $n$ circles, and its fundamental group is also $F_n$. Since both our graph and the wedge sum of finite circles (essentially also a connected graph, with loops) are CW complex $K(G,1)$ that have the same fundamental group, they are homotopy equivalent.
answered Dec 1 '18 at 5:09
chikurinchikurin
899
edited Dec 1 '18 at 5:23
answered Dec 1 '18 at 5:09
chikurinchikurin
899
answered Dec 1 '18 at 5:09
chikurinchikurin
899
answered Dec 1 '18 at 5:09
chikurinchikurin
899
899
$begingroup$
This is a bit high-powered given the likely context of what OP knows.
$endgroup$
– Joshua Mundinger
Dec 1 '18 at 5:26
$begingroup$
Probably yes... I was just searching for similar questions and kind of want to write an answer for myself or if by any chance it can help the community.
$endgroup$
– chikurin
Dec 1 '18 at 5:33
add a comment |
$begingroup$
This is a bit high-powered given the likely context of what OP knows.
$endgroup$
– Joshua Mundinger
Dec 1 '18 at 5:26
$begingroup$
Probably yes... I was just searching for similar questions and kind of want to write an answer for myself or if by any chance it can help the community.
$endgroup$
– chikurin
Dec 1 '18 at 5:33
$begingroup$
This is a bit high-powered given the likely context of what OP knows.
$endgroup$
– Joshua Mundinger
Dec 1 '18 at 5:26
$begingroup$
This is a bit high-powered given the likely context of what OP knows.
$endgroup$
– Joshua Mundinger
Dec 1 '18 at 5:26
$begingroup$
Probably yes... I was just searching for similar questions and kind of want to write an answer for myself or if by any chance it can help the community.
$endgroup$
– chikurin
Dec 1 '18 at 5:33
$begingroup$
Probably yes... I was just searching for similar questions and kind of want to write an answer for myself or if by any chance it can help the community.
$endgroup$
– chikurin
Dec 1 '18 at 5:33
add a comment |
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$begingroup$
You have to show that contracting an edge $Gto G/e$ is a homotopy equivalence. This can be done by hand, and is surprisingly tricky. You need to define maps in both directions (in the definition of homotopy equivalence), and defining the map from $G/eto G$ is the tricky bit but if you think about it, there aren't that many ways to do it. You could also appeal to more general theorems that state that contracting contractible subspaces yield homotopy equivalences under favorable circumstances. It depends what theorems you have access to.
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 5:24
$begingroup$
See math.stackexchange.com/questions/21705/…
$endgroup$
– Cheerful Parsnip
Dec 1 '18 at 7:26