Mixing
Simplicial homology and homeomorphisms
In Hatcher's book, in the introduction page of singular homology, he mentions that "it is obvious that homeomorphic spaces have the same singular homology, in contrast to simplicial homology". However I thought that this was also true for simplicial homology (and looking at the construction I don't see why this would not be true for simplicial homology).
What does he mean by this statement? Does he point at the fact that not all spaces are triangulable and hence do not admit the construction of simplicial homology?
algebraic-topology
asked Sep 17 '18 at 16:09
NDewolfNDewolf
538210
add a comment |
In Hatcher's book, in the introduction page of singular homology, he mentions that "it is obvious that homeomorphic spaces have the same singular homology, in contrast to simplicial homology". However I thought that this was also true for simplicial homology (and looking at the construction I don't see why this would not be true for simplicial homology).
What does he mean by this statement? Does he point at the fact that not all spaces are triangulable and hence do not admit the construction of simplicial homology?
algebraic-topology
asked Sep 17 '18 at 16:09
NDewolfNDewolf
538210
5
He means that for simplicial homology it's not obvious, because you may have different triangulations for a given space
– Max
Sep 17 '18 at 16:13
1
Rather a triangulable space will have many different triangulations, and it is not obvious that each of these triangulations will have isomorphic homology.
– Lord Shark the Unknown
Sep 17 '18 at 16:14
add a comment |
In Hatcher's book, in the introduction page of singular homology, he mentions that "it is obvious that homeomorphic spaces have the same singular homology, in contrast to simplicial homology". However I thought that this was also true for simplicial homology (and looking at the construction I don't see why this would not be true for simplicial homology).
What does he mean by this statement? Does he point at the fact that not all spaces are triangulable and hence do not admit the construction of simplicial homology?
algebraic-topology
asked Sep 17 '18 at 16:09
NDewolfNDewolf
538210
In Hatcher's book, in the introduction page of singular homology, he mentions that "it is obvious that homeomorphic spaces have the same singular homology, in contrast to simplicial homology". However I thought that this was also true for simplicial homology (and looking at the construction I don't see why this would not be true for simplicial homology).
What does he mean by this statement? Does he point at the fact that not all spaces are triangulable and hence do not admit the construction of simplicial homology?
algebraic-topology
algebraic-topology
asked Sep 17 '18 at 16:09
NDewolfNDewolf
538210
asked Sep 17 '18 at 16:09
NDewolfNDewolf
538210
asked Sep 17 '18 at 16:09
NDewolfNDewolf
538210
asked Sep 17 '18 at 16:09
NDewolfNDewolf
538210
asked Sep 17 '18 at 16:09
NDewolfNDewolf
538210
538210
5
He means that for simplicial homology it's not obvious, because you may have different triangulations for a given space
– Max
Sep 17 '18 at 16:13
1
Rather a triangulable space will have many different triangulations, and it is not obvious that each of these triangulations will have isomorphic homology.
– Lord Shark the Unknown
Sep 17 '18 at 16:14
add a comment |
5
He means that for simplicial homology it's not obvious, because you may have different triangulations for a given space
– Max
Sep 17 '18 at 16:13
1
Rather a triangulable space will have many different triangulations, and it is not obvious that each of these triangulations will have isomorphic homology.
– Lord Shark the Unknown
Sep 17 '18 at 16:14
5
5
He means that for simplicial homology it's not obvious, because you may have different triangulations for a given space
– Max
Sep 17 '18 at 16:13
He means that for simplicial homology it's not obvious, because you may have different triangulations for a given space
– Max
Sep 17 '18 at 16:13
1
1
Rather a triangulable space will have many different triangulations, and it is not obvious that each of these triangulations will have isomorphic homology.
– Lord Shark the Unknown
Sep 17 '18 at 16:14
Rather a triangulable space will have many different triangulations, and it is not obvious that each of these triangulations will have isomorphic homology.
– Lord Shark the Unknown
Sep 17 '18 at 16:14
add a comment |
2 Answers
2
active
oldest
votes
As Max points out in the comments, the same space might have different triangulations.
To elaborate, the construction of singular homology relies only on continuous maps, so it's "obvious" that singular homology is invariant under homeomorphism. However, the construction of simplicial homology relies on the additional structure of a triangulation of your space. You could have multiple triangulations of a space which might -- conceivably! -- produce different homology.
Because of this dependence in the construction of simplicial homology, you need to prove that simplicial homology is independent of triangulation. That extra work is why Hatcher says it is not obvious that simplicial homology is a homeomorphism invariant.
answered Sep 17 '18 at 16:32
NealNeal
23.5k23784
So in this spirit it is only obvious that two homeomorphic spaces have the same simplicial homology with respect to a certain triangulation (which is how i looked at it)?
– NDewolf
Sep 17 '18 at 17:04
@NDewolf Not quite; if two triangulated spaces have a homeomorphism between them that respects the triangulation, then it is "obvious" that they have isomorphic simplicial homology. Key is that you need to take triangulations into account.
– Neal
Sep 17 '18 at 18:59
@NDewolf ... the important point here is that simplicial homology is a priori constructed on a tuple $(mbox{topological space}, mbox{triangulation})$. Even when you only consider one space $X$, you have to do work to show that simplicial homology doesn't change when you change the triangulation. Why should $H_*(X,mathcal{T}) sim H_*(X,mathcal{S})$ for two different triangulations $mathcal{T}$, $mathcal{S}$?
– Neal
Sep 17 '18 at 19:07
But isn't a triangulated space just a topological space homeomorphic to a simplicial complex. So given a triangulated space $(X, mathcal{T})$ and a space $Y$ homeomorphic to $X$, isn't $Y$ trivially also triangulated by $mathcal{T}$? (Just by composing the homeomorphisms)
– NDewolf
Sep 17 '18 at 21:12
@NDewolf yes- but what if $Y$ has a different triangulation? No guarantee the homology constructed from that same triangulation is the same as the homology constructed from the pushed-forward triangulation.
– Neal
Sep 17 '18 at 21:33
|
show 1 more comment
Neal's answer explains the problem of different triangulations. Nevertheless it turns out that the simplicial homology $H_*(mathcal{T})$, where $mathcal{T}$ is a simplicial complex triangulating $X$, is a topological invariant of $X$. That is, if $X_1, X_2$ are homeomorphic and $mathcal{T}_i$ are triangulations of $X_i$, then $H_*(mathcal{T}_1) approx H_*(mathcal{T}_2)$. The standard proof relies on "identifying" the simplicial homology of a triangulation of $X$ with the singular homology of $X$. This is a genuine topological proof.
Historically, the need for a topological proof was not so obviuos. In the "early days" mathematicians conjectured that a combinatorial proof was possible. It is a simple observation that if $mathcal{T}$ is a triangulation of $X$ and $mathcal{T}'$ is a subdivision of $mathcal{T}$, then $H_*(mathcal{T}) approx H_*(mathcal{T}')$. Now the so-called Hauptvermutung said that any two triangulations of a triangulable space have a common subdivision. This would obviuosly prove that $H_*(mathcal{T})$ is a topological invariant of $X$.
Unfortunately the Hauptvermutung fails as was shown by John Milnor in 1961. Therefore there is no combinatorial proof.
See for example https://en.wikipedia.org/wiki/Hauptvermutung.
answered Nov 21 '18 at 19:24
Paul FrostPaul Frost
9,5002631
add a comment |
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2 Answers
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As Max points out in the comments, the same space might have different triangulations.
To elaborate, the construction of singular homology relies only on continuous maps, so it's "obvious" that singular homology is invariant under homeomorphism. However, the construction of simplicial homology relies on the additional structure of a triangulation of your space. You could have multiple triangulations of a space which might -- conceivably! -- produce different homology.
Because of this dependence in the construction of simplicial homology, you need to prove that simplicial homology is independent of triangulation. That extra work is why Hatcher says it is not obvious that simplicial homology is a homeomorphism invariant.
answered Sep 17 '18 at 16:32
NealNeal
23.5k23784
So in this spirit it is only obvious that two homeomorphic spaces have the same simplicial homology with respect to a certain triangulation (which is how i looked at it)?
– NDewolf
Sep 17 '18 at 17:04
@NDewolf Not quite; if two triangulated spaces have a homeomorphism between them that respects the triangulation, then it is "obvious" that they have isomorphic simplicial homology. Key is that you need to take triangulations into account.
– Neal
Sep 17 '18 at 18:59
@NDewolf ... the important point here is that simplicial homology is a priori constructed on a tuple $(mbox{topological space}, mbox{triangulation})$. Even when you only consider one space $X$, you have to do work to show that simplicial homology doesn't change when you change the triangulation. Why should $H_*(X,mathcal{T}) sim H_*(X,mathcal{S})$ for two different triangulations $mathcal{T}$, $mathcal{S}$?
– Neal
Sep 17 '18 at 19:07
But isn't a triangulated space just a topological space homeomorphic to a simplicial complex. So given a triangulated space $(X, mathcal{T})$ and a space $Y$ homeomorphic to $X$, isn't $Y$ trivially also triangulated by $mathcal{T}$? (Just by composing the homeomorphisms)
– NDewolf
Sep 17 '18 at 21:12
@NDewolf yes- but what if $Y$ has a different triangulation? No guarantee the homology constructed from that same triangulation is the same as the homology constructed from the pushed-forward triangulation.
– Neal
Sep 17 '18 at 21:33
|
show 1 more comment
As Max points out in the comments, the same space might have different triangulations.
To elaborate, the construction of singular homology relies only on continuous maps, so it's "obvious" that singular homology is invariant under homeomorphism. However, the construction of simplicial homology relies on the additional structure of a triangulation of your space. You could have multiple triangulations of a space which might -- conceivably! -- produce different homology.
Because of this dependence in the construction of simplicial homology, you need to prove that simplicial homology is independent of triangulation. That extra work is why Hatcher says it is not obvious that simplicial homology is a homeomorphism invariant.
answered Sep 17 '18 at 16:32
NealNeal
23.5k23784
So in this spirit it is only obvious that two homeomorphic spaces have the same simplicial homology with respect to a certain triangulation (which is how i looked at it)?
– NDewolf
Sep 17 '18 at 17:04
@NDewolf Not quite; if two triangulated spaces have a homeomorphism between them that respects the triangulation, then it is "obvious" that they have isomorphic simplicial homology. Key is that you need to take triangulations into account.
– Neal
Sep 17 '18 at 18:59
@NDewolf ... the important point here is that simplicial homology is a priori constructed on a tuple $(mbox{topological space}, mbox{triangulation})$. Even when you only consider one space $X$, you have to do work to show that simplicial homology doesn't change when you change the triangulation. Why should $H_*(X,mathcal{T}) sim H_*(X,mathcal{S})$ for two different triangulations $mathcal{T}$, $mathcal{S}$?
– Neal
Sep 17 '18 at 19:07
But isn't a triangulated space just a topological space homeomorphic to a simplicial complex. So given a triangulated space $(X, mathcal{T})$ and a space $Y$ homeomorphic to $X$, isn't $Y$ trivially also triangulated by $mathcal{T}$? (Just by composing the homeomorphisms)
– NDewolf
Sep 17 '18 at 21:12
@NDewolf yes- but what if $Y$ has a different triangulation? No guarantee the homology constructed from that same triangulation is the same as the homology constructed from the pushed-forward triangulation.
– Neal
Sep 17 '18 at 21:33
|
show 1 more comment
As Max points out in the comments, the same space might have different triangulations.
To elaborate, the construction of singular homology relies only on continuous maps, so it's "obvious" that singular homology is invariant under homeomorphism. However, the construction of simplicial homology relies on the additional structure of a triangulation of your space. You could have multiple triangulations of a space which might -- conceivably! -- produce different homology.
Because of this dependence in the construction of simplicial homology, you need to prove that simplicial homology is independent of triangulation. That extra work is why Hatcher says it is not obvious that simplicial homology is a homeomorphism invariant.
answered Sep 17 '18 at 16:32
NealNeal
23.5k23784
As Max points out in the comments, the same space might have different triangulations.
To elaborate, the construction of singular homology relies only on continuous maps, so it's "obvious" that singular homology is invariant under homeomorphism. However, the construction of simplicial homology relies on the additional structure of a triangulation of your space. You could have multiple triangulations of a space which might -- conceivably! -- produce different homology.
Because of this dependence in the construction of simplicial homology, you need to prove that simplicial homology is independent of triangulation. That extra work is why Hatcher says it is not obvious that simplicial homology is a homeomorphism invariant.
answered Sep 17 '18 at 16:32
NealNeal
23.5k23784
answered Sep 17 '18 at 16:32
NealNeal
23.5k23784
answered Sep 17 '18 at 16:32
NealNeal
23.5k23784
answered Sep 17 '18 at 16:32
NealNeal
23.5k23784
23.5k23784
So in this spirit it is only obvious that two homeomorphic spaces have the same simplicial homology with respect to a certain triangulation (which is how i looked at it)?
– NDewolf
Sep 17 '18 at 17:04
@NDewolf Not quite; if two triangulated spaces have a homeomorphism between them that respects the triangulation, then it is "obvious" that they have isomorphic simplicial homology. Key is that you need to take triangulations into account.
– Neal
Sep 17 '18 at 18:59
@NDewolf ... the important point here is that simplicial homology is a priori constructed on a tuple $(mbox{topological space}, mbox{triangulation})$. Even when you only consider one space $X$, you have to do work to show that simplicial homology doesn't change when you change the triangulation. Why should $H_*(X,mathcal{T}) sim H_*(X,mathcal{S})$ for two different triangulations $mathcal{T}$, $mathcal{S}$?
– Neal
Sep 17 '18 at 19:07
But isn't a triangulated space just a topological space homeomorphic to a simplicial complex. So given a triangulated space $(X, mathcal{T})$ and a space $Y$ homeomorphic to $X$, isn't $Y$ trivially also triangulated by $mathcal{T}$? (Just by composing the homeomorphisms)
– NDewolf
Sep 17 '18 at 21:12
@NDewolf yes- but what if $Y$ has a different triangulation? No guarantee the homology constructed from that same triangulation is the same as the homology constructed from the pushed-forward triangulation.
– Neal
Sep 17 '18 at 21:33
|
show 1 more comment
So in this spirit it is only obvious that two homeomorphic spaces have the same simplicial homology with respect to a certain triangulation (which is how i looked at it)?
– NDewolf
Sep 17 '18 at 17:04
@NDewolf Not quite; if two triangulated spaces have a homeomorphism between them that respects the triangulation, then it is "obvious" that they have isomorphic simplicial homology. Key is that you need to take triangulations into account.
– Neal
Sep 17 '18 at 18:59
@NDewolf ... the important point here is that simplicial homology is a priori constructed on a tuple $(mbox{topological space}, mbox{triangulation})$. Even when you only consider one space $X$, you have to do work to show that simplicial homology doesn't change when you change the triangulation. Why should $H_*(X,mathcal{T}) sim H_*(X,mathcal{S})$ for two different triangulations $mathcal{T}$, $mathcal{S}$?
– Neal
Sep 17 '18 at 19:07
But isn't a triangulated space just a topological space homeomorphic to a simplicial complex. So given a triangulated space $(X, mathcal{T})$ and a space $Y$ homeomorphic to $X$, isn't $Y$ trivially also triangulated by $mathcal{T}$? (Just by composing the homeomorphisms)
– NDewolf
Sep 17 '18 at 21:12
@NDewolf yes- but what if $Y$ has a different triangulation? No guarantee the homology constructed from that same triangulation is the same as the homology constructed from the pushed-forward triangulation.
– Neal
Sep 17 '18 at 21:33
So in this spirit it is only obvious that two homeomorphic spaces have the same simplicial homology with respect to a certain triangulation (which is how i looked at it)?
– NDewolf
Sep 17 '18 at 17:04
So in this spirit it is only obvious that two homeomorphic spaces have the same simplicial homology with respect to a certain triangulation (which is how i looked at it)?
– NDewolf
Sep 17 '18 at 17:04
@NDewolf Not quite; if two triangulated spaces have a homeomorphism between them that respects the triangulation, then it is "obvious" that they have isomorphic simplicial homology. Key is that you need to take triangulations into account.
– Neal
Sep 17 '18 at 18:59
@NDewolf Not quite; if two triangulated spaces have a homeomorphism between them that respects the triangulation, then it is "obvious" that they have isomorphic simplicial homology. Key is that you need to take triangulations into account.
– Neal
Sep 17 '18 at 18:59
@NDewolf ... the important point here is that simplicial homology is a priori constructed on a tuple $(mbox{topological space}, mbox{triangulation})$. Even when you only consider one space $X$, you have to do work to show that simplicial homology doesn't change when you change the triangulation. Why should $H_*(X,mathcal{T}) sim H_*(X,mathcal{S})$ for two different triangulations $mathcal{T}$, $mathcal{S}$?
– Neal
Sep 17 '18 at 19:07
@NDewolf ... the important point here is that simplicial homology is a priori constructed on a tuple $(mbox{topological space}, mbox{triangulation})$. Even when you only consider one space $X$, you have to do work to show that simplicial homology doesn't change when you change the triangulation. Why should $H_*(X,mathcal{T}) sim H_*(X,mathcal{S})$ for two different triangulations $mathcal{T}$, $mathcal{S}$?
– Neal
Sep 17 '18 at 19:07
But isn't a triangulated space just a topological space homeomorphic to a simplicial complex. So given a triangulated space $(X, mathcal{T})$ and a space $Y$ homeomorphic to $X$, isn't $Y$ trivially also triangulated by $mathcal{T}$? (Just by composing the homeomorphisms)
– NDewolf
Sep 17 '18 at 21:12
But isn't a triangulated space just a topological space homeomorphic to a simplicial complex. So given a triangulated space $(X, mathcal{T})$ and a space $Y$ homeomorphic to $X$, isn't $Y$ trivially also triangulated by $mathcal{T}$? (Just by composing the homeomorphisms)
– NDewolf
Sep 17 '18 at 21:12
@NDewolf yes- but what if $Y$ has a different triangulation? No guarantee the homology constructed from that same triangulation is the same as the homology constructed from the pushed-forward triangulation.
– Neal
Sep 17 '18 at 21:33
@NDewolf yes- but what if $Y$ has a different triangulation? No guarantee the homology constructed from that same triangulation is the same as the homology constructed from the pushed-forward triangulation.
– Neal
Sep 17 '18 at 21:33
|
show 1 more comment
Neal's answer explains the problem of different triangulations. Nevertheless it turns out that the simplicial homology $H_*(mathcal{T})$, where $mathcal{T}$ is a simplicial complex triangulating $X$, is a topological invariant of $X$. That is, if $X_1, X_2$ are homeomorphic and $mathcal{T}_i$ are triangulations of $X_i$, then $H_*(mathcal{T}_1) approx H_*(mathcal{T}_2)$. The standard proof relies on "identifying" the simplicial homology of a triangulation of $X$ with the singular homology of $X$. This is a genuine topological proof.
Historically, the need for a topological proof was not so obviuos. In the "early days" mathematicians conjectured that a combinatorial proof was possible. It is a simple observation that if $mathcal{T}$ is a triangulation of $X$ and $mathcal{T}'$ is a subdivision of $mathcal{T}$, then $H_*(mathcal{T}) approx H_*(mathcal{T}')$. Now the so-called Hauptvermutung said that any two triangulations of a triangulable space have a common subdivision. This would obviuosly prove that $H_*(mathcal{T})$ is a topological invariant of $X$.
Unfortunately the Hauptvermutung fails as was shown by John Milnor in 1961. Therefore there is no combinatorial proof.
See for example https://en.wikipedia.org/wiki/Hauptvermutung.
answered Nov 21 '18 at 19:24
Paul FrostPaul Frost
9,5002631
add a comment |
Neal's answer explains the problem of different triangulations. Nevertheless it turns out that the simplicial homology $H_*(mathcal{T})$, where $mathcal{T}$ is a simplicial complex triangulating $X$, is a topological invariant of $X$. That is, if $X_1, X_2$ are homeomorphic and $mathcal{T}_i$ are triangulations of $X_i$, then $H_*(mathcal{T}_1) approx H_*(mathcal{T}_2)$. The standard proof relies on "identifying" the simplicial homology of a triangulation of $X$ with the singular homology of $X$. This is a genuine topological proof.
Historically, the need for a topological proof was not so obviuos. In the "early days" mathematicians conjectured that a combinatorial proof was possible. It is a simple observation that if $mathcal{T}$ is a triangulation of $X$ and $mathcal{T}'$ is a subdivision of $mathcal{T}$, then $H_*(mathcal{T}) approx H_*(mathcal{T}')$. Now the so-called Hauptvermutung said that any two triangulations of a triangulable space have a common subdivision. This would obviuosly prove that $H_*(mathcal{T})$ is a topological invariant of $X$.
Unfortunately the Hauptvermutung fails as was shown by John Milnor in 1961. Therefore there is no combinatorial proof.
See for example https://en.wikipedia.org/wiki/Hauptvermutung.
answered Nov 21 '18 at 19:24
Paul FrostPaul Frost
9,5002631
add a comment |
Neal's answer explains the problem of different triangulations. Nevertheless it turns out that the simplicial homology $H_*(mathcal{T})$, where $mathcal{T}$ is a simplicial complex triangulating $X$, is a topological invariant of $X$. That is, if $X_1, X_2$ are homeomorphic and $mathcal{T}_i$ are triangulations of $X_i$, then $H_*(mathcal{T}_1) approx H_*(mathcal{T}_2)$. The standard proof relies on "identifying" the simplicial homology of a triangulation of $X$ with the singular homology of $X$. This is a genuine topological proof.
Historically, the need for a topological proof was not so obviuos. In the "early days" mathematicians conjectured that a combinatorial proof was possible. It is a simple observation that if $mathcal{T}$ is a triangulation of $X$ and $mathcal{T}'$ is a subdivision of $mathcal{T}$, then $H_*(mathcal{T}) approx H_*(mathcal{T}')$. Now the so-called Hauptvermutung said that any two triangulations of a triangulable space have a common subdivision. This would obviuosly prove that $H_*(mathcal{T})$ is a topological invariant of $X$.
Unfortunately the Hauptvermutung fails as was shown by John Milnor in 1961. Therefore there is no combinatorial proof.
See for example https://en.wikipedia.org/wiki/Hauptvermutung.
answered Nov 21 '18 at 19:24
Paul FrostPaul Frost
9,5002631
Neal's answer explains the problem of different triangulations. Nevertheless it turns out that the simplicial homology $H_*(mathcal{T})$, where $mathcal{T}$ is a simplicial complex triangulating $X$, is a topological invariant of $X$. That is, if $X_1, X_2$ are homeomorphic and $mathcal{T}_i$ are triangulations of $X_i$, then $H_*(mathcal{T}_1) approx H_*(mathcal{T}_2)$. The standard proof relies on "identifying" the simplicial homology of a triangulation of $X$ with the singular homology of $X$. This is a genuine topological proof.
Historically, the need for a topological proof was not so obviuos. In the "early days" mathematicians conjectured that a combinatorial proof was possible. It is a simple observation that if $mathcal{T}$ is a triangulation of $X$ and $mathcal{T}'$ is a subdivision of $mathcal{T}$, then $H_*(mathcal{T}) approx H_*(mathcal{T}')$. Now the so-called Hauptvermutung said that any two triangulations of a triangulable space have a common subdivision. This would obviuosly prove that $H_*(mathcal{T})$ is a topological invariant of $X$.
Unfortunately the Hauptvermutung fails as was shown by John Milnor in 1961. Therefore there is no combinatorial proof.
See for example https://en.wikipedia.org/wiki/Hauptvermutung.
answered Nov 21 '18 at 19:24
Paul FrostPaul Frost
9,5002631
answered Nov 21 '18 at 19:24
Paul FrostPaul Frost
9,5002631
answered Nov 21 '18 at 19:24
Paul FrostPaul Frost
9,5002631
answered Nov 21 '18 at 19:24
Paul FrostPaul Frost
9,5002631
9,5002631
add a comment |
add a comment |
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5
He means that for simplicial homology it's not obvious, because you may have different triangulations for a given space
– Max
Sep 17 '18 at 16:13
1
Rather a triangulable space will have many different triangulations, and it is not obvious that each of these triangulations will have isomorphic homology.
– Lord Shark the Unknown
Sep 17 '18 at 16:14