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







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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

















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







share|cite|improve this question
























  • 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















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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 10:51









Christoph

11.6k1541




11.6k1541










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




















  • 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












1 Answer
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1
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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.






share|cite|improve this answer





















  • 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











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1 Answer
1






active

oldest

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active

oldest

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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.






share|cite|improve this answer





















  • 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















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.






share|cite|improve this answer





















  • 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













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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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