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Singular homology - take every possible map?
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I am reading singular homology in Hatcher. Let $X$ be a topological space. We let $C_n(X)$ be generated by singular $n$-simplicies $sigma:Delta^nto X$.
What I don't understand is, what collection of $n$-simplicies do we want? Do we just take every single possible distinct map $Delta^nto X$? Say I am mapping $3$-simplicies into the space $X=Delta^3$, do I not have uncountably many choices of maps?
I can see in this way why the point space has a single singular $n$-cell for each $n$, since we must map each simplex to the point, if we are taking every possible map.
algebraic-topology homology-cohomology
asked Jul 28 at 22:56
Heaven Decays
263
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up vote
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favorite
I am reading singular homology in Hatcher. Let $X$ be a topological space. We let $C_n(X)$ be generated by singular $n$-simplicies $sigma:Delta^nto X$.
What I don't understand is, what collection of $n$-simplicies do we want? Do we just take every single possible distinct map $Delta^nto X$? Say I am mapping $3$-simplicies into the space $X=Delta^3$, do I not have uncountably many choices of maps?
I can see in this way why the point space has a single singular $n$-cell for each $n$, since we must map each simplex to the point, if we are taking every possible map.
algebraic-topology homology-cohomology
asked Jul 28 at 22:56
Heaven Decays
263
5
Yes, all such continuous maps. It will be extraordinarily many. An equivalence relation will take care of that.
– Randall
Jul 28 at 22:58
@Randall What is the equivalence relation? Or you mean when we take homology it doesn't matter, since we identify the faces of the simplicies with the standard simplex?
– Heaven Decays
Jul 28 at 23:01
Yes, forming the homology groups.
– Randall
Jul 28 at 23:01
@Randall In some sense it seems easy to work out what the maps in the kernel of the boundary map are, at least for small $n$. What are we really identifying when we quotient the image though?
– Heaven Decays
Jul 28 at 23:59
@Randall I guess I should formulate my question more carefully, in a distinct MSE question.
– Heaven Decays
Jul 29 at 0:06
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am reading singular homology in Hatcher. Let $X$ be a topological space. We let $C_n(X)$ be generated by singular $n$-simplicies $sigma:Delta^nto X$.
What I don't understand is, what collection of $n$-simplicies do we want? Do we just take every single possible distinct map $Delta^nto X$? Say I am mapping $3$-simplicies into the space $X=Delta^3$, do I not have uncountably many choices of maps?
I can see in this way why the point space has a single singular $n$-cell for each $n$, since we must map each simplex to the point, if we are taking every possible map.
algebraic-topology homology-cohomology
asked Jul 28 at 22:56
Heaven Decays
263
I am reading singular homology in Hatcher. Let $X$ be a topological space. We let $C_n(X)$ be generated by singular $n$-simplicies $sigma:Delta^nto X$.
What I don't understand is, what collection of $n$-simplicies do we want? Do we just take every single possible distinct map $Delta^nto X$? Say I am mapping $3$-simplicies into the space $X=Delta^3$, do I not have uncountably many choices of maps?
I can see in this way why the point space has a single singular $n$-cell for each $n$, since we must map each simplex to the point, if we are taking every possible map.
algebraic-topology homology-cohomology
asked Jul 28 at 22:56
Heaven Decays
263
asked Jul 28 at 22:56
Heaven Decays
263
asked Jul 28 at 22:56
Heaven Decays
263
asked Jul 28 at 22:56
Heaven Decays
263
263
5
Yes, all such continuous maps. It will be extraordinarily many. An equivalence relation will take care of that.
– Randall
Jul 28 at 22:58
@Randall What is the equivalence relation? Or you mean when we take homology it doesn't matter, since we identify the faces of the simplicies with the standard simplex?
– Heaven Decays
Jul 28 at 23:01
Yes, forming the homology groups.
– Randall
Jul 28 at 23:01
@Randall In some sense it seems easy to work out what the maps in the kernel of the boundary map are, at least for small $n$. What are we really identifying when we quotient the image though?
– Heaven Decays
Jul 28 at 23:59
@Randall I guess I should formulate my question more carefully, in a distinct MSE question.
– Heaven Decays
Jul 29 at 0:06
add a comment |Â
5
Yes, all such continuous maps. It will be extraordinarily many. An equivalence relation will take care of that.
– Randall
Jul 28 at 22:58
@Randall What is the equivalence relation? Or you mean when we take homology it doesn't matter, since we identify the faces of the simplicies with the standard simplex?
– Heaven Decays
Jul 28 at 23:01
Yes, forming the homology groups.
– Randall
Jul 28 at 23:01
@Randall In some sense it seems easy to work out what the maps in the kernel of the boundary map are, at least for small $n$. What are we really identifying when we quotient the image though?
– Heaven Decays
Jul 28 at 23:59
@Randall I guess I should formulate my question more carefully, in a distinct MSE question.
– Heaven Decays
Jul 29 at 0:06
5
5
Yes, all such continuous maps. It will be extraordinarily many. An equivalence relation will take care of that.
– Randall
Jul 28 at 22:58
Yes, all such continuous maps. It will be extraordinarily many. An equivalence relation will take care of that.
– Randall
Jul 28 at 22:58
@Randall What is the equivalence relation? Or you mean when we take homology it doesn't matter, since we identify the faces of the simplicies with the standard simplex?
– Heaven Decays
Jul 28 at 23:01
@Randall What is the equivalence relation? Or you mean when we take homology it doesn't matter, since we identify the faces of the simplicies with the standard simplex?
– Heaven Decays
Jul 28 at 23:01
Yes, forming the homology groups.
– Randall
Jul 28 at 23:01
Yes, forming the homology groups.
– Randall
Jul 28 at 23:01
@Randall In some sense it seems easy to work out what the maps in the kernel of the boundary map are, at least for small $n$. What are we really identifying when we quotient the image though?
– Heaven Decays
Jul 28 at 23:59
@Randall In some sense it seems easy to work out what the maps in the kernel of the boundary map are, at least for small $n$. What are we really identifying when we quotient the image though?
– Heaven Decays
Jul 28 at 23:59
@Randall I guess I should formulate my question more carefully, in a distinct MSE question.
– Heaven Decays
Jul 29 at 0:06
@Randall I guess I should formulate my question more carefully, in a distinct MSE question.
– Heaven Decays
Jul 29 at 0:06
add a comment |Â
1 Answer
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Historically homology groups were first defined for simplicial complexes which are purely combinatorial entities. See e.g. https://en.wikipedia.org/wiki/Simplicial_complex and https://en.wikipedia.org/wiki/Simplicial_homology.
Let us consider your example $Delta^3$. We obtain one generator of the non-oriented simplicial chain complex in dimension $3$ resp. two generators of the oriented simplicial chain complex in dimension $3$. This is a very effective method to associate chain complexes and homology groups to simplicial complexes.
What is the relationship between topological spaces and simplicial complexes? Each simplicial complex $K$ has a geometric realization $lvert K rvert$ which is a topological space consisting of Euclidean simplices which are convex hulls of points in general position in some $mathbbR^n$ (each combinatorial $n$-simplex which is a finite set of vertices is realized as a Euclidean simplex, and these Euclidean simplices are glued together by the corresponding combinatorial incidences of the faces). But certainly not every space is the geometric realization of simplicial complex. Moreover, two non-isomorphic simplicial complexes may have homeomorphic geometric realizations - are their homology groups isomorphic ("topological invariance of simplicial homology")? It was discovered that this question cannot be answered by purely combinatorial methods. This was the origin of singular homology. A singular simplex in a space $X$ is any map $sigma : Delta^n to X$. In general this produces uncountably many singular simplices (even for the simplest spaces). The resulting chain complexes are very big, but if you go to homology groups the size massively collapses. In fact, the singular homology groups of the geometric realization $lvert K rvert $ of a simplicial complex $K$ are isomorphic to the simplicial homology groups of $K$. This proves the topological invariance of simplicial homology and provides a method to effectively compute the homology groups of polyhedra (spaces $X$ which have triangulations, i.e. admit a homeomorphism $h ; X to lvert K rvert$ where $K$ is simplicial complex).
Singular homology is excellent because it is conceptually simple. It is defined for any space without referring to additional structural components (which are usually not uniquely determined by the space itself, see e.g. https://en.wikipedia.org/wiki/Hauptvermutung), even the beginner will easily understand it and topological invariance is obvious from the definition (that is why we take all singular simplices and don't choose a "smaller" assortment). However, it is practically impossible to compute singular homology groups based on the definition of the singular chain complex. But singular homology satisfies the Eilenberg–Steenrod axioms https://en.wikipedia.org/wiki/Eilenberg%E2%80%93Steenrod_axioms and we can compute a lot of homology groups using only these axioms. See any book on algebraic topology. The Eilenberg–Steenrod axioms determine homology theories on finite CW-complexes uniquely (up to natural isomorphism) by their coefficient groups $H_0(ast)$ (where $ast$ is a one-point space). In case of the standard singular homology theory we have $H_0(ast) approx mathbbZ$ which is one of the few cases where an effective computation is possible.
My opinion is that the main purpose of singular homology is to establish the existence of a homology theory which satisfies the Eilenberg–Steenrod axioms and is defined for all topological spaces. There are a lot of other constructions, but they are not simpler and some are only defined on certain classes of spaces. The material distinctions between the various constructions are unveiled only on spaces which are no finite CW-cmplexes, and then it becomes apparent that there are homology theories which perform better then singular homology.
answered Jul 29 at 0:47
Paul Frost
3,593420
This makes sense. I guess my concern then is that $C_n(X)$ is usually not a finitely generated object, and the kernel and image of the boundary map will then have to be non-finitely generated. How am I then meant to compute singular homology, even for a circle. Here it's super easy using simplicial homology, where as you say these coincide anyway. Is singular more to give the theory, than to use to compute?
– Heaven Decays
Jul 29 at 1:15
True that the chain groups will not be finitely generated, but that will not stop the quotient group from being nice and f.g.
– Randall
Jul 29 at 1:23
And, I find singular homology to be wonderful for giving the theory, and I also prefer it computationally, though I admit other more combinatorial models are often simpler. Singular theory just makes more sense to me.
– Randall
Jul 29 at 1:23
1
Not to hijack this, but I also prefer the singular theory because it works for all spaces with no simplicial/CW/manifold/whatever assumptions.
– Randall
Jul 29 at 1:25
1
For some more comments on the history, I like to refer to M. Barr, "Oriented singular homology" tac.mta.ca/tac/volumes/1995/n1/1-01abs.html .
– Ronnie Brown
Jul 29 at 11:32
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Historically homology groups were first defined for simplicial complexes which are purely combinatorial entities. See e.g. https://en.wikipedia.org/wiki/Simplicial_complex and https://en.wikipedia.org/wiki/Simplicial_homology.
Let us consider your example $Delta^3$. We obtain one generator of the non-oriented simplicial chain complex in dimension $3$ resp. two generators of the oriented simplicial chain complex in dimension $3$. This is a very effective method to associate chain complexes and homology groups to simplicial complexes.
What is the relationship between topological spaces and simplicial complexes? Each simplicial complex $K$ has a geometric realization $lvert K rvert$ which is a topological space consisting of Euclidean simplices which are convex hulls of points in general position in some $mathbbR^n$ (each combinatorial $n$-simplex which is a finite set of vertices is realized as a Euclidean simplex, and these Euclidean simplices are glued together by the corresponding combinatorial incidences of the faces). But certainly not every space is the geometric realization of simplicial complex. Moreover, two non-isomorphic simplicial complexes may have homeomorphic geometric realizations - are their homology groups isomorphic ("topological invariance of simplicial homology")? It was discovered that this question cannot be answered by purely combinatorial methods. This was the origin of singular homology. A singular simplex in a space $X$ is any map $sigma : Delta^n to X$. In general this produces uncountably many singular simplices (even for the simplest spaces). The resulting chain complexes are very big, but if you go to homology groups the size massively collapses. In fact, the singular homology groups of the geometric realization $lvert K rvert $ of a simplicial complex $K$ are isomorphic to the simplicial homology groups of $K$. This proves the topological invariance of simplicial homology and provides a method to effectively compute the homology groups of polyhedra (spaces $X$ which have triangulations, i.e. admit a homeomorphism $h ; X to lvert K rvert$ where $K$ is simplicial complex).
Singular homology is excellent because it is conceptually simple. It is defined for any space without referring to additional structural components (which are usually not uniquely determined by the space itself, see e.g. https://en.wikipedia.org/wiki/Hauptvermutung), even the beginner will easily understand it and topological invariance is obvious from the definition (that is why we take all singular simplices and don't choose a "smaller" assortment). However, it is practically impossible to compute singular homology groups based on the definition of the singular chain complex. But singular homology satisfies the Eilenberg–Steenrod axioms https://en.wikipedia.org/wiki/Eilenberg%E2%80%93Steenrod_axioms and we can compute a lot of homology groups using only these axioms. See any book on algebraic topology. The Eilenberg–Steenrod axioms determine homology theories on finite CW-complexes uniquely (up to natural isomorphism) by their coefficient groups $H_0(ast)$ (where $ast$ is a one-point space). In case of the standard singular homology theory we have $H_0(ast) approx mathbbZ$ which is one of the few cases where an effective computation is possible.
My opinion is that the main purpose of singular homology is to establish the existence of a homology theory which satisfies the Eilenberg–Steenrod axioms and is defined for all topological spaces. There are a lot of other constructions, but they are not simpler and some are only defined on certain classes of spaces. The material distinctions between the various constructions are unveiled only on spaces which are no finite CW-cmplexes, and then it becomes apparent that there are homology theories which perform better then singular homology.
answered Jul 29 at 0:47
Paul Frost
3,593420
This makes sense. I guess my concern then is that $C_n(X)$ is usually not a finitely generated object, and the kernel and image of the boundary map will then have to be non-finitely generated. How am I then meant to compute singular homology, even for a circle. Here it's super easy using simplicial homology, where as you say these coincide anyway. Is singular more to give the theory, than to use to compute?
– Heaven Decays
Jul 29 at 1:15
True that the chain groups will not be finitely generated, but that will not stop the quotient group from being nice and f.g.
– Randall
Jul 29 at 1:23
And, I find singular homology to be wonderful for giving the theory, and I also prefer it computationally, though I admit other more combinatorial models are often simpler. Singular theory just makes more sense to me.
– Randall
Jul 29 at 1:23
1
Not to hijack this, but I also prefer the singular theory because it works for all spaces with no simplicial/CW/manifold/whatever assumptions.
– Randall
Jul 29 at 1:25
1
For some more comments on the history, I like to refer to M. Barr, "Oriented singular homology" tac.mta.ca/tac/volumes/1995/n1/1-01abs.html .
– Ronnie Brown
Jul 29 at 11:32
 |Â
show 4 more comments
up vote
2
down vote
Historically homology groups were first defined for simplicial complexes which are purely combinatorial entities. See e.g. https://en.wikipedia.org/wiki/Simplicial_complex and https://en.wikipedia.org/wiki/Simplicial_homology.
Let us consider your example $Delta^3$. We obtain one generator of the non-oriented simplicial chain complex in dimension $3$ resp. two generators of the oriented simplicial chain complex in dimension $3$. This is a very effective method to associate chain complexes and homology groups to simplicial complexes.
What is the relationship between topological spaces and simplicial complexes? Each simplicial complex $K$ has a geometric realization $lvert K rvert$ which is a topological space consisting of Euclidean simplices which are convex hulls of points in general position in some $mathbbR^n$ (each combinatorial $n$-simplex which is a finite set of vertices is realized as a Euclidean simplex, and these Euclidean simplices are glued together by the corresponding combinatorial incidences of the faces). But certainly not every space is the geometric realization of simplicial complex. Moreover, two non-isomorphic simplicial complexes may have homeomorphic geometric realizations - are their homology groups isomorphic ("topological invariance of simplicial homology")? It was discovered that this question cannot be answered by purely combinatorial methods. This was the origin of singular homology. A singular simplex in a space $X$ is any map $sigma : Delta^n to X$. In general this produces uncountably many singular simplices (even for the simplest spaces). The resulting chain complexes are very big, but if you go to homology groups the size massively collapses. In fact, the singular homology groups of the geometric realization $lvert K rvert $ of a simplicial complex $K$ are isomorphic to the simplicial homology groups of $K$. This proves the topological invariance of simplicial homology and provides a method to effectively compute the homology groups of polyhedra (spaces $X$ which have triangulations, i.e. admit a homeomorphism $h ; X to lvert K rvert$ where $K$ is simplicial complex).
Singular homology is excellent because it is conceptually simple. It is defined for any space without referring to additional structural components (which are usually not uniquely determined by the space itself, see e.g. https://en.wikipedia.org/wiki/Hauptvermutung), even the beginner will easily understand it and topological invariance is obvious from the definition (that is why we take all singular simplices and don't choose a "smaller" assortment). However, it is practically impossible to compute singular homology groups based on the definition of the singular chain complex. But singular homology satisfies the Eilenberg–Steenrod axioms https://en.wikipedia.org/wiki/Eilenberg%E2%80%93Steenrod_axioms and we can compute a lot of homology groups using only these axioms. See any book on algebraic topology. The Eilenberg–Steenrod axioms determine homology theories on finite CW-complexes uniquely (up to natural isomorphism) by their coefficient groups $H_0(ast)$ (where $ast$ is a one-point space). In case of the standard singular homology theory we have $H_0(ast) approx mathbbZ$ which is one of the few cases where an effective computation is possible.
My opinion is that the main purpose of singular homology is to establish the existence of a homology theory which satisfies the Eilenberg–Steenrod axioms and is defined for all topological spaces. There are a lot of other constructions, but they are not simpler and some are only defined on certain classes of spaces. The material distinctions between the various constructions are unveiled only on spaces which are no finite CW-cmplexes, and then it becomes apparent that there are homology theories which perform better then singular homology.
answered Jul 29 at 0:47
Paul Frost
3,593420
This makes sense. I guess my concern then is that $C_n(X)$ is usually not a finitely generated object, and the kernel and image of the boundary map will then have to be non-finitely generated. How am I then meant to compute singular homology, even for a circle. Here it's super easy using simplicial homology, where as you say these coincide anyway. Is singular more to give the theory, than to use to compute?
– Heaven Decays
Jul 29 at 1:15
True that the chain groups will not be finitely generated, but that will not stop the quotient group from being nice and f.g.
– Randall
Jul 29 at 1:23
And, I find singular homology to be wonderful for giving the theory, and I also prefer it computationally, though I admit other more combinatorial models are often simpler. Singular theory just makes more sense to me.
– Randall
Jul 29 at 1:23
1
Not to hijack this, but I also prefer the singular theory because it works for all spaces with no simplicial/CW/manifold/whatever assumptions.
– Randall
Jul 29 at 1:25
1
For some more comments on the history, I like to refer to M. Barr, "Oriented singular homology" tac.mta.ca/tac/volumes/1995/n1/1-01abs.html .
– Ronnie Brown
Jul 29 at 11:32
 |Â
show 4 more comments
up vote
2
down vote
up vote
2
down vote
Historically homology groups were first defined for simplicial complexes which are purely combinatorial entities. See e.g. https://en.wikipedia.org/wiki/Simplicial_complex and https://en.wikipedia.org/wiki/Simplicial_homology.
Let us consider your example $Delta^3$. We obtain one generator of the non-oriented simplicial chain complex in dimension $3$ resp. two generators of the oriented simplicial chain complex in dimension $3$. This is a very effective method to associate chain complexes and homology groups to simplicial complexes.
What is the relationship between topological spaces and simplicial complexes? Each simplicial complex $K$ has a geometric realization $lvert K rvert$ which is a topological space consisting of Euclidean simplices which are convex hulls of points in general position in some $mathbbR^n$ (each combinatorial $n$-simplex which is a finite set of vertices is realized as a Euclidean simplex, and these Euclidean simplices are glued together by the corresponding combinatorial incidences of the faces). But certainly not every space is the geometric realization of simplicial complex. Moreover, two non-isomorphic simplicial complexes may have homeomorphic geometric realizations - are their homology groups isomorphic ("topological invariance of simplicial homology")? It was discovered that this question cannot be answered by purely combinatorial methods. This was the origin of singular homology. A singular simplex in a space $X$ is any map $sigma : Delta^n to X$. In general this produces uncountably many singular simplices (even for the simplest spaces). The resulting chain complexes are very big, but if you go to homology groups the size massively collapses. In fact, the singular homology groups of the geometric realization $lvert K rvert $ of a simplicial complex $K$ are isomorphic to the simplicial homology groups of $K$. This proves the topological invariance of simplicial homology and provides a method to effectively compute the homology groups of polyhedra (spaces $X$ which have triangulations, i.e. admit a homeomorphism $h ; X to lvert K rvert$ where $K$ is simplicial complex).
Singular homology is excellent because it is conceptually simple. It is defined for any space without referring to additional structural components (which are usually not uniquely determined by the space itself, see e.g. https://en.wikipedia.org/wiki/Hauptvermutung), even the beginner will easily understand it and topological invariance is obvious from the definition (that is why we take all singular simplices and don't choose a "smaller" assortment). However, it is practically impossible to compute singular homology groups based on the definition of the singular chain complex. But singular homology satisfies the Eilenberg–Steenrod axioms https://en.wikipedia.org/wiki/Eilenberg%E2%80%93Steenrod_axioms and we can compute a lot of homology groups using only these axioms. See any book on algebraic topology. The Eilenberg–Steenrod axioms determine homology theories on finite CW-complexes uniquely (up to natural isomorphism) by their coefficient groups $H_0(ast)$ (where $ast$ is a one-point space). In case of the standard singular homology theory we have $H_0(ast) approx mathbbZ$ which is one of the few cases where an effective computation is possible.
My opinion is that the main purpose of singular homology is to establish the existence of a homology theory which satisfies the Eilenberg–Steenrod axioms and is defined for all topological spaces. There are a lot of other constructions, but they are not simpler and some are only defined on certain classes of spaces. The material distinctions between the various constructions are unveiled only on spaces which are no finite CW-cmplexes, and then it becomes apparent that there are homology theories which perform better then singular homology.
answered Jul 29 at 0:47
Paul Frost
3,593420
Historically homology groups were first defined for simplicial complexes which are purely combinatorial entities. See e.g. https://en.wikipedia.org/wiki/Simplicial_complex and https://en.wikipedia.org/wiki/Simplicial_homology.
Let us consider your example $Delta^3$. We obtain one generator of the non-oriented simplicial chain complex in dimension $3$ resp. two generators of the oriented simplicial chain complex in dimension $3$. This is a very effective method to associate chain complexes and homology groups to simplicial complexes.
What is the relationship between topological spaces and simplicial complexes? Each simplicial complex $K$ has a geometric realization $lvert K rvert$ which is a topological space consisting of Euclidean simplices which are convex hulls of points in general position in some $mathbbR^n$ (each combinatorial $n$-simplex which is a finite set of vertices is realized as a Euclidean simplex, and these Euclidean simplices are glued together by the corresponding combinatorial incidences of the faces). But certainly not every space is the geometric realization of simplicial complex. Moreover, two non-isomorphic simplicial complexes may have homeomorphic geometric realizations - are their homology groups isomorphic ("topological invariance of simplicial homology")? It was discovered that this question cannot be answered by purely combinatorial methods. This was the origin of singular homology. A singular simplex in a space $X$ is any map $sigma : Delta^n to X$. In general this produces uncountably many singular simplices (even for the simplest spaces). The resulting chain complexes are very big, but if you go to homology groups the size massively collapses. In fact, the singular homology groups of the geometric realization $lvert K rvert $ of a simplicial complex $K$ are isomorphic to the simplicial homology groups of $K$. This proves the topological invariance of simplicial homology and provides a method to effectively compute the homology groups of polyhedra (spaces $X$ which have triangulations, i.e. admit a homeomorphism $h ; X to lvert K rvert$ where $K$ is simplicial complex).
Singular homology is excellent because it is conceptually simple. It is defined for any space without referring to additional structural components (which are usually not uniquely determined by the space itself, see e.g. https://en.wikipedia.org/wiki/Hauptvermutung), even the beginner will easily understand it and topological invariance is obvious from the definition (that is why we take all singular simplices and don't choose a "smaller" assortment). However, it is practically impossible to compute singular homology groups based on the definition of the singular chain complex. But singular homology satisfies the Eilenberg–Steenrod axioms https://en.wikipedia.org/wiki/Eilenberg%E2%80%93Steenrod_axioms and we can compute a lot of homology groups using only these axioms. See any book on algebraic topology. The Eilenberg–Steenrod axioms determine homology theories on finite CW-complexes uniquely (up to natural isomorphism) by their coefficient groups $H_0(ast)$ (where $ast$ is a one-point space). In case of the standard singular homology theory we have $H_0(ast) approx mathbbZ$ which is one of the few cases where an effective computation is possible.
My opinion is that the main purpose of singular homology is to establish the existence of a homology theory which satisfies the Eilenberg–Steenrod axioms and is defined for all topological spaces. There are a lot of other constructions, but they are not simpler and some are only defined on certain classes of spaces. The material distinctions between the various constructions are unveiled only on spaces which are no finite CW-cmplexes, and then it becomes apparent that there are homology theories which perform better then singular homology.
answered Jul 29 at 0:47
Paul Frost
3,593420
edited Jul 29 at 8:21
answered Jul 29 at 0:47
Paul Frost
3,593420
answered Jul 29 at 0:47
Paul Frost
3,593420
answered Jul 29 at 0:47
Paul Frost
3,593420
3,593420
This makes sense. I guess my concern then is that $C_n(X)$ is usually not a finitely generated object, and the kernel and image of the boundary map will then have to be non-finitely generated. How am I then meant to compute singular homology, even for a circle. Here it's super easy using simplicial homology, where as you say these coincide anyway. Is singular more to give the theory, than to use to compute?
– Heaven Decays
Jul 29 at 1:15
True that the chain groups will not be finitely generated, but that will not stop the quotient group from being nice and f.g.
– Randall
Jul 29 at 1:23
And, I find singular homology to be wonderful for giving the theory, and I also prefer it computationally, though I admit other more combinatorial models are often simpler. Singular theory just makes more sense to me.
– Randall
Jul 29 at 1:23
1
Not to hijack this, but I also prefer the singular theory because it works for all spaces with no simplicial/CW/manifold/whatever assumptions.
– Randall
Jul 29 at 1:25
1
For some more comments on the history, I like to refer to M. Barr, "Oriented singular homology" tac.mta.ca/tac/volumes/1995/n1/1-01abs.html .
– Ronnie Brown
Jul 29 at 11:32
 |Â
show 4 more comments
This makes sense. I guess my concern then is that $C_n(X)$ is usually not a finitely generated object, and the kernel and image of the boundary map will then have to be non-finitely generated. How am I then meant to compute singular homology, even for a circle. Here it's super easy using simplicial homology, where as you say these coincide anyway. Is singular more to give the theory, than to use to compute?
– Heaven Decays
Jul 29 at 1:15
True that the chain groups will not be finitely generated, but that will not stop the quotient group from being nice and f.g.
– Randall
Jul 29 at 1:23
And, I find singular homology to be wonderful for giving the theory, and I also prefer it computationally, though I admit other more combinatorial models are often simpler. Singular theory just makes more sense to me.
– Randall
Jul 29 at 1:23
1
Not to hijack this, but I also prefer the singular theory because it works for all spaces with no simplicial/CW/manifold/whatever assumptions.
– Randall
Jul 29 at 1:25
1
For some more comments on the history, I like to refer to M. Barr, "Oriented singular homology" tac.mta.ca/tac/volumes/1995/n1/1-01abs.html .
– Ronnie Brown
Jul 29 at 11:32
This makes sense. I guess my concern then is that $C_n(X)$ is usually not a finitely generated object, and the kernel and image of the boundary map will then have to be non-finitely generated. How am I then meant to compute singular homology, even for a circle. Here it's super easy using simplicial homology, where as you say these coincide anyway. Is singular more to give the theory, than to use to compute?
– Heaven Decays
Jul 29 at 1:15
This makes sense. I guess my concern then is that $C_n(X)$ is usually not a finitely generated object, and the kernel and image of the boundary map will then have to be non-finitely generated. How am I then meant to compute singular homology, even for a circle. Here it's super easy using simplicial homology, where as you say these coincide anyway. Is singular more to give the theory, than to use to compute?
– Heaven Decays
Jul 29 at 1:15
True that the chain groups will not be finitely generated, but that will not stop the quotient group from being nice and f.g.
– Randall
Jul 29 at 1:23
True that the chain groups will not be finitely generated, but that will not stop the quotient group from being nice and f.g.
– Randall
Jul 29 at 1:23
And, I find singular homology to be wonderful for giving the theory, and I also prefer it computationally, though I admit other more combinatorial models are often simpler. Singular theory just makes more sense to me.
– Randall
Jul 29 at 1:23
And, I find singular homology to be wonderful for giving the theory, and I also prefer it computationally, though I admit other more combinatorial models are often simpler. Singular theory just makes more sense to me.
– Randall
Jul 29 at 1:23
1
1
Not to hijack this, but I also prefer the singular theory because it works for all spaces with no simplicial/CW/manifold/whatever assumptions.
– Randall
Jul 29 at 1:25
Not to hijack this, but I also prefer the singular theory because it works for all spaces with no simplicial/CW/manifold/whatever assumptions.
– Randall
Jul 29 at 1:25
1
1
For some more comments on the history, I like to refer to M. Barr, "Oriented singular homology" tac.mta.ca/tac/volumes/1995/n1/1-01abs.html .
– Ronnie Brown
Jul 29 at 11:32
For some more comments on the history, I like to refer to M. Barr, "Oriented singular homology" tac.mta.ca/tac/volumes/1995/n1/1-01abs.html .
– Ronnie Brown
Jul 29 at 11:32
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5
Yes, all such continuous maps. It will be extraordinarily many. An equivalence relation will take care of that.
– Randall
Jul 28 at 22:58
@Randall What is the equivalence relation? Or you mean when we take homology it doesn't matter, since we identify the faces of the simplicies with the standard simplex?
– Heaven Decays
Jul 28 at 23:01
Yes, forming the homology groups.
– Randall
Jul 28 at 23:01
@Randall In some sense it seems easy to work out what the maps in the kernel of the boundary map are, at least for small $n$. What are we really identifying when we quotient the image though?
– Heaven Decays
Jul 28 at 23:59
@Randall I guess I should formulate my question more carefully, in a distinct MSE question.
– Heaven Decays
Jul 29 at 0:06