chain homotopy equivalence and quasi-isomorphism
2
$begingroup$
Suppose $(C,d)$ and $(D,delta)$ are two chain complexes over a field and $f:Cto D$ is a chain map.
- We say $f$ is a quasi-isomorphism if it induces an isomorphism of the homology groups $H(C,d)to H(D,delta)$.
- We say $f$ is a chain homotopy equivalence if there is a chain map $g:Dto C$ so that $gf$ and $fg$ are chain homotopic to the identity maps.
It seems that the item 2 is stronger and can imply the item 1, but my question is that, conversely, under what additional conditions the item 1 can also imply the item 2? Thank you.
algebraic-topology homology-cohomology homological-algebra homotopy-theory
asked Dec 26 '18 at 22:41
HangHang
502415
$endgroup$
add a comment |
2
$begingroup$
Suppose $(C,d)$ and $(D,delta)$ are two chain complexes over a field and $f:Cto D$ is a chain map.
- We say $f$ is a quasi-isomorphism if it induces an isomorphism of the homology groups $H(C,d)to H(D,delta)$.
- We say $f$ is a chain homotopy equivalence if there is a chain map $g:Dto C$ so that $gf$ and $fg$ are chain homotopic to the identity maps.
It seems that the item 2 is stronger and can imply the item 1, but my question is that, conversely, under what additional conditions the item 1 can also imply the item 2? Thank you.
algebraic-topology homology-cohomology homological-algebra homotopy-theory
asked Dec 26 '18 at 22:41
HangHang
502415
$endgroup$
1
$begingroup$
mathoverflow.net/questions/59390/… (the question statement even!) looks like it answers this somewhat? I'd like to see the full detail of that though, personally.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:19
$begingroup$
Thank you! That post states that this is true over a field. Do you know why? It seems that this is sort of standard facts. Actually I do not need very advanced stuffs.
$endgroup$
– Hang
Dec 27 '18 at 0:25
$begingroup$
So this is because any chain complex over a field is split, then exercise 1.4.4 of Weibel (aix1.uottawa.ca/~rblute/COURSE2/weibel.pdf) is to show that split implies chain homotopic to its homology. Hopefully reading section 1.4 of Weibel has all the details needed to prove this! Once you have that each complex is chain homotopic to its own homology, any quasi isomorphism will give you a chain homotopy by going to the homology, using the isomorphism, then back.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:42
$begingroup$
@AlexJBest Thank you very much!
$endgroup$
– Hang
Dec 27 '18 at 0:43
add a comment |
2
2
2
$begingroup$
Suppose $(C,d)$ and $(D,delta)$ are two chain complexes over a field and $f:Cto D$ is a chain map.
- We say $f$ is a quasi-isomorphism if it induces an isomorphism of the homology groups $H(C,d)to H(D,delta)$.
- We say $f$ is a chain homotopy equivalence if there is a chain map $g:Dto C$ so that $gf$ and $fg$ are chain homotopic to the identity maps.
It seems that the item 2 is stronger and can imply the item 1, but my question is that, conversely, under what additional conditions the item 1 can also imply the item 2? Thank you.
algebraic-topology homology-cohomology homological-algebra homotopy-theory
asked Dec 26 '18 at 22:41
HangHang
502415
$endgroup$
Suppose $(C,d)$ and $(D,delta)$ are two chain complexes over a field and $f:Cto D$ is a chain map.
- We say $f$ is a quasi-isomorphism if it induces an isomorphism of the homology groups $H(C,d)to H(D,delta)$.
- We say $f$ is a chain homotopy equivalence if there is a chain map $g:Dto C$ so that $gf$ and $fg$ are chain homotopic to the identity maps.
It seems that the item 2 is stronger and can imply the item 1, but my question is that, conversely, under what additional conditions the item 1 can also imply the item 2? Thank you.
algebraic-topology homology-cohomology homological-algebra homotopy-theory
algebraic-topology homology-cohomology homological-algebra homotopy-theory
asked Dec 26 '18 at 22:41
HangHang
502415
asked Dec 26 '18 at 22:41
HangHang
502415
asked Dec 26 '18 at 22:41
HangHang
502415
asked Dec 26 '18 at 22:41
HangHang
502415
asked Dec 26 '18 at 22:41
HangHang
502415
502415
1
$begingroup$
mathoverflow.net/questions/59390/… (the question statement even!) looks like it answers this somewhat? I'd like to see the full detail of that though, personally.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:19
$begingroup$
Thank you! That post states that this is true over a field. Do you know why? It seems that this is sort of standard facts. Actually I do not need very advanced stuffs.
$endgroup$
– Hang
Dec 27 '18 at 0:25
$begingroup$
So this is because any chain complex over a field is split, then exercise 1.4.4 of Weibel (aix1.uottawa.ca/~rblute/COURSE2/weibel.pdf) is to show that split implies chain homotopic to its homology. Hopefully reading section 1.4 of Weibel has all the details needed to prove this! Once you have that each complex is chain homotopic to its own homology, any quasi isomorphism will give you a chain homotopy by going to the homology, using the isomorphism, then back.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:42
$begingroup$
@AlexJBest Thank you very much!
$endgroup$
– Hang
Dec 27 '18 at 0:43
add a comment |
1
$begingroup$
mathoverflow.net/questions/59390/… (the question statement even!) looks like it answers this somewhat? I'd like to see the full detail of that though, personally.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:19
$begingroup$
Thank you! That post states that this is true over a field. Do you know why? It seems that this is sort of standard facts. Actually I do not need very advanced stuffs.
$endgroup$
– Hang
Dec 27 '18 at 0:25
$begingroup$
So this is because any chain complex over a field is split, then exercise 1.4.4 of Weibel (aix1.uottawa.ca/~rblute/COURSE2/weibel.pdf) is to show that split implies chain homotopic to its homology. Hopefully reading section 1.4 of Weibel has all the details needed to prove this! Once you have that each complex is chain homotopic to its own homology, any quasi isomorphism will give you a chain homotopy by going to the homology, using the isomorphism, then back.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:42
$begingroup$
@AlexJBest Thank you very much!
$endgroup$
– Hang
Dec 27 '18 at 0:43
1
1
$begingroup$
mathoverflow.net/questions/59390/… (the question statement even!) looks like it answers this somewhat? I'd like to see the full detail of that though, personally.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:19
$begingroup$
mathoverflow.net/questions/59390/… (the question statement even!) looks like it answers this somewhat? I'd like to see the full detail of that though, personally.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:19
$begingroup$
Thank you! That post states that this is true over a field. Do you know why? It seems that this is sort of standard facts. Actually I do not need very advanced stuffs.
$endgroup$
– Hang
Dec 27 '18 at 0:25
$begingroup$
Thank you! That post states that this is true over a field. Do you know why? It seems that this is sort of standard facts. Actually I do not need very advanced stuffs.
$endgroup$
– Hang
Dec 27 '18 at 0:25
$begingroup$
So this is because any chain complex over a field is split, then exercise 1.4.4 of Weibel (aix1.uottawa.ca/~rblute/COURSE2/weibel.pdf) is to show that split implies chain homotopic to its homology. Hopefully reading section 1.4 of Weibel has all the details needed to prove this! Once you have that each complex is chain homotopic to its own homology, any quasi isomorphism will give you a chain homotopy by going to the homology, using the isomorphism, then back.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:42
$begingroup$
So this is because any chain complex over a field is split, then exercise 1.4.4 of Weibel (aix1.uottawa.ca/~rblute/COURSE2/weibel.pdf) is to show that split implies chain homotopic to its homology. Hopefully reading section 1.4 of Weibel has all the details needed to prove this! Once you have that each complex is chain homotopic to its own homology, any quasi isomorphism will give you a chain homotopy by going to the homology, using the isomorphism, then back.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:42
$begingroup$
@AlexJBest Thank you very much!
$endgroup$
– Hang
Dec 27 '18 at 0:43
$begingroup$
@AlexJBest Thank you very much!
$endgroup$
– Hang
Dec 27 '18 at 0:43
add a comment |
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1
$begingroup$
mathoverflow.net/questions/59390/… (the question statement even!) looks like it answers this somewhat? I'd like to see the full detail of that though, personally.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:19
$begingroup$
Thank you! That post states that this is true over a field. Do you know why? It seems that this is sort of standard facts. Actually I do not need very advanced stuffs.
$endgroup$
– Hang
Dec 27 '18 at 0:25
$begingroup$
So this is because any chain complex over a field is split, then exercise 1.4.4 of Weibel (aix1.uottawa.ca/~rblute/COURSE2/weibel.pdf) is to show that split implies chain homotopic to its homology. Hopefully reading section 1.4 of Weibel has all the details needed to prove this! Once you have that each complex is chain homotopic to its own homology, any quasi isomorphism will give you a chain homotopy by going to the homology, using the isomorphism, then back.
$endgroup$
– Alex J Best
Dec 27 '18 at 0:42
$begingroup$
@AlexJBest Thank you very much!
$endgroup$
– Hang
Dec 27 '18 at 0:43