Singular Homology: every $0$-chain is a $0$-circle
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I am having trouble understanding this fact, that is deemed as trivial and thus not proved in most books.
First of all, I understand that the boundary operator $operatorname{Bdy}$ goes from $S_n$ to $S_{n-1}$ (where $S_n$ is the group of $n$-chains).
In this case, which is the codomain? I assumed it is $S_0$ as well, exceptionally.
Secondly, if $f$ is a $0$-chain, it should have the $0$-simplex, a point, as it's domain.
$operatorname{Bdy}(f)$ should be $f(0)$ in this case. But $0$ does $0$ belong to the $0$-simplex?
algebraic-topology homology-cohomology simplex
edited Nov 22 at 10:51
Christoph
11.6k1541
asked Nov 22 at 10:44
Francesco Bilotta
82
add a comment |
up vote
1
down vote
favorite
I am having trouble understanding this fact, that is deemed as trivial and thus not proved in most books.
First of all, I understand that the boundary operator $operatorname{Bdy}$ goes from $S_n$ to $S_{n-1}$ (where $S_n$ is the group of $n$-chains).
In this case, which is the codomain? I assumed it is $S_0$ as well, exceptionally.
Secondly, if $f$ is a $0$-chain, it should have the $0$-simplex, a point, as it's domain.
$operatorname{Bdy}(f)$ should be $f(0)$ in this case. But $0$ does $0$ belong to the $0$-simplex?
algebraic-topology homology-cohomology simplex
edited Nov 22 at 10:51
Christoph
11.6k1541
asked Nov 22 at 10:44
Francesco Bilotta
82
The boundary map $partial_0$ either goes $S_0to 0$ (and you get the usual homology) or it goes $S_0to mathbb Z$ sending each $0$-simplex to $1$, you then get reduced homology.
– Christoph
Nov 22 at 10:48
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am having trouble understanding this fact, that is deemed as trivial and thus not proved in most books.
First of all, I understand that the boundary operator $operatorname{Bdy}$ goes from $S_n$ to $S_{n-1}$ (where $S_n$ is the group of $n$-chains).
In this case, which is the codomain? I assumed it is $S_0$ as well, exceptionally.
Secondly, if $f$ is a $0$-chain, it should have the $0$-simplex, a point, as it's domain.
$operatorname{Bdy}(f)$ should be $f(0)$ in this case. But $0$ does $0$ belong to the $0$-simplex?
algebraic-topology homology-cohomology simplex
edited Nov 22 at 10:51
Christoph
11.6k1541
asked Nov 22 at 10:44
Francesco Bilotta
82
I am having trouble understanding this fact, that is deemed as trivial and thus not proved in most books.
First of all, I understand that the boundary operator $operatorname{Bdy}$ goes from $S_n$ to $S_{n-1}$ (where $S_n$ is the group of $n$-chains).
In this case, which is the codomain? I assumed it is $S_0$ as well, exceptionally.
Secondly, if $f$ is a $0$-chain, it should have the $0$-simplex, a point, as it's domain.
$operatorname{Bdy}(f)$ should be $f(0)$ in this case. But $0$ does $0$ belong to the $0$-simplex?
algebraic-topology homology-cohomology simplex
algebraic-topology homology-cohomology simplex
edited Nov 22 at 10:51
Christoph
11.6k1541
asked Nov 22 at 10:44
Francesco Bilotta
82
edited Nov 22 at 10:51
Christoph
11.6k1541
asked Nov 22 at 10:44
Francesco Bilotta
82
edited Nov 22 at 10:51
Christoph
11.6k1541
edited Nov 22 at 10:51
Christoph
11.6k1541
edited Nov 22 at 10:51
Christoph
11.6k1541
11.6k1541
asked Nov 22 at 10:44
Francesco Bilotta
82
asked Nov 22 at 10:44
Francesco Bilotta
82
asked Nov 22 at 10:44
Francesco Bilotta
82
82
The boundary map $partial_0$ either goes $S_0to 0$ (and you get the usual homology) or it goes $S_0to mathbb Z$ sending each $0$-simplex to $1$, you then get reduced homology.
– Christoph
Nov 22 at 10:48
add a comment |
The boundary map $partial_0$ either goes $S_0to 0$ (and you get the usual homology) or it goes $S_0to mathbb Z$ sending each $0$-simplex to $1$, you then get reduced homology.
– Christoph
Nov 22 at 10:48
The boundary map $partial_0$ either goes $S_0to 0$ (and you get the usual homology) or it goes $S_0to mathbb Z$ sending each $0$-simplex to $1$, you then get reduced homology.
– Christoph
Nov 22 at 10:48
The boundary map $partial_0$ either goes $S_0to 0$ (and you get the usual homology) or it goes $S_0to mathbb Z$ sending each $0$-simplex to $1$, you then get reduced homology.
– Christoph
Nov 22 at 10:48
add a comment |
1 Answer
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up vote
1
down vote
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The codomain is $0$ literally, this also makes every $0$-chain into a $0$-cycle(respectively, if you like to think about it that way: a point has no boundary, if you think about it, every point is a degenerate simplex with trivial boundary).
I know this is quite a technicality, however, it helps a lot although sometimes it is also useful to pass to the reduced homology, where you kill the class of the basepoint, since precisely this technicality can make stuff weird.
answered Nov 22 at 10:51
Enkidu
1,02618
Note that $0$ as the codomain here and in my comment is used to denote the trivial group with only one element.
– Christoph
Nov 22 at 10:53
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The codomain is $0$ literally, this also makes every $0$-chain into a $0$-cycle(respectively, if you like to think about it that way: a point has no boundary, if you think about it, every point is a degenerate simplex with trivial boundary).
I know this is quite a technicality, however, it helps a lot although sometimes it is also useful to pass to the reduced homology, where you kill the class of the basepoint, since precisely this technicality can make stuff weird.
answered Nov 22 at 10:51
Enkidu
1,02618
Note that $0$ as the codomain here and in my comment is used to denote the trivial group with only one element.
– Christoph
Nov 22 at 10:53
add a comment |
up vote
1
down vote
accepted
The codomain is $0$ literally, this also makes every $0$-chain into a $0$-cycle(respectively, if you like to think about it that way: a point has no boundary, if you think about it, every point is a degenerate simplex with trivial boundary).
I know this is quite a technicality, however, it helps a lot although sometimes it is also useful to pass to the reduced homology, where you kill the class of the basepoint, since precisely this technicality can make stuff weird.
answered Nov 22 at 10:51
Enkidu
1,02618
Note that $0$ as the codomain here and in my comment is used to denote the trivial group with only one element.
– Christoph
Nov 22 at 10:53
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The codomain is $0$ literally, this also makes every $0$-chain into a $0$-cycle(respectively, if you like to think about it that way: a point has no boundary, if you think about it, every point is a degenerate simplex with trivial boundary).
I know this is quite a technicality, however, it helps a lot although sometimes it is also useful to pass to the reduced homology, where you kill the class of the basepoint, since precisely this technicality can make stuff weird.
answered Nov 22 at 10:51
Enkidu
1,02618
The codomain is $0$ literally, this also makes every $0$-chain into a $0$-cycle(respectively, if you like to think about it that way: a point has no boundary, if you think about it, every point is a degenerate simplex with trivial boundary).
I know this is quite a technicality, however, it helps a lot although sometimes it is also useful to pass to the reduced homology, where you kill the class of the basepoint, since precisely this technicality can make stuff weird.
answered Nov 22 at 10:51
Enkidu
1,02618
answered Nov 22 at 10:51
Enkidu
1,02618
answered Nov 22 at 10:51
Enkidu
1,02618
answered Nov 22 at 10:51
Enkidu
1,02618
1,02618
Note that $0$ as the codomain here and in my comment is used to denote the trivial group with only one element.
– Christoph
Nov 22 at 10:53
add a comment |
Note that $0$ as the codomain here and in my comment is used to denote the trivial group with only one element.
– Christoph
Nov 22 at 10:53
Note that $0$ as the codomain here and in my comment is used to denote the trivial group with only one element.
– Christoph
Nov 22 at 10:53
Note that $0$ as the codomain here and in my comment is used to denote the trivial group with only one element.
– Christoph
Nov 22 at 10:53
add a comment |
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The boundary map $partial_0$ either goes $S_0to 0$ (and you get the usual homology) or it goes $S_0to mathbb Z$ sending each $0$-simplex to $1$, you then get reduced homology.
– Christoph
Nov 22 at 10:48