Mixing
Cohomology groups $Coker(i_{n-1}^*)$ depends on only $H_{n-1}(C)$ and $G$. Hatcher
$begingroup$
On page 193 of Hatcher at the bottom. He says
$$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 (vi)$$
Note that the group $Coker(i_{n-1}^*)$ that we are interested in is
$H^1(F;G)$ where $F$ is the free resolution in (vi). Part (b) of the following lemma therefore
shows that $Coker i∗_{n−1}$ depends only on $H_{n−1}(C)$ and $G$.
Which two resolution is he using to derive this result?
Is it $(iv)$ and $(v)$?? Very confused.
How does he pass through $H^n(C;G)$?
algebraic-topology proof-explanation
asked Jan 27 at 5:26
HawkHawk
5,5551140109
$endgroup$
add a comment |
$begingroup$
On page 193 of Hatcher at the bottom. He says
$$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 (vi)$$
Note that the group $Coker(i_{n-1}^*)$ that we are interested in is
$H^1(F;G)$ where $F$ is the free resolution in (vi). Part (b) of the following lemma therefore
shows that $Coker i∗_{n−1}$ depends only on $H_{n−1}(C)$ and $G$.
Which two resolution is he using to derive this result?
Is it $(iv)$ and $(v)$?? Very confused.
How does he pass through $H^n(C;G)$?
algebraic-topology proof-explanation
asked Jan 27 at 5:26
HawkHawk
5,5551140109
$endgroup$
add a comment |
$begingroup$
On page 193 of Hatcher at the bottom. He says
$$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 (vi)$$
Note that the group $Coker(i_{n-1}^*)$ that we are interested in is
$H^1(F;G)$ where $F$ is the free resolution in (vi). Part (b) of the following lemma therefore
shows that $Coker i∗_{n−1}$ depends only on $H_{n−1}(C)$ and $G$.
Which two resolution is he using to derive this result?
Is it $(iv)$ and $(v)$?? Very confused.
How does he pass through $H^n(C;G)$?
algebraic-topology proof-explanation
asked Jan 27 at 5:26
HawkHawk
5,5551140109
$endgroup$
On page 193 of Hatcher at the bottom. He says
$$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 (vi)$$
Note that the group $Coker(i_{n-1}^*)$ that we are interested in is
$H^1(F;G)$ where $F$ is the free resolution in (vi). Part (b) of the following lemma therefore
shows that $Coker i∗_{n−1}$ depends only on $H_{n−1}(C)$ and $G$.
Which two resolution is he using to derive this result?
Is it $(iv)$ and $(v)$?? Very confused.
How does he pass through $H^n(C;G)$?
algebraic-topology proof-explanation
algebraic-topology proof-explanation
asked Jan 27 at 5:26
HawkHawk
5,5551140109
asked Jan 27 at 5:26
HawkHawk
5,5551140109
asked Jan 27 at 5:26
HawkHawk
5,5551140109
asked Jan 27 at 5:26
HawkHawk
5,5551140109
asked Jan 27 at 5:26
HawkHawk
5,5551140109
5,5551140109
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Math: the claim is that if we change the chain complex $C$ and pick a totally different complex C', requiring only that it have the same homology in the $n-1$ spot, then the complex $text{Hom}(C, G)$ and the complex $text{Hom}(C', G)$ would still have the same $text{cok}(iota^*_{n-1})$ (up to canonical isomorphism). The claim is attained by taking the sequence (vi) and the analogous sequence (vi') for $C'$ and applying part two of the lemma.
Philosophy: if this is too much arrow-ception you might try skipping these two pages and just reading the statement of the universal coefficient theorem and doing a couple exercises, then coming back to it. It's one of those theorems that you kind of have to prove for yourself instead of following someone else's proof because it's 8 million steps in a row, any individual one of which isn't too bad, but all of which together are confusing the first 7 to 12 times.
answered Jan 27 at 5:50
hunterhunter
15.3k32640
$endgroup$
$begingroup$
No I know what the lemma is saying. I am asking which specific sequence they are talking about. If you are taking (vi), what is the (vi')? This doesn't exist, it is artificial.
$endgroup$
– Hawk
Jan 27 at 6:09
$begingroup$
(vi') is (vi) but for $C'$ instead of $C$. That is what it means to say that the cokernel doesn't depend on the complex but only the homology.
$endgroup$
– hunter
Jan 27 at 6:18
$begingroup$
So you are telling me we are putting the same sequence by itself and all the vertical maps are identity...?
$endgroup$
– Hawk
Jan 27 at 6:25
$begingroup$
@Hawk I am saying we are taking two different resolutions of $H_{n-1}(C)$. the vertical maps aren't the identity (except the third vertical map, which is).
$endgroup$
– hunter
Jan 27 at 6:27
$begingroup$
$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 $ and this mysterious notational $0 to B_{n-1}' to Z_{n-1}' to H_{n-1}(C') to 0 $ row by row...? How do we know $C to C'$ is an isomorphism (or I guess identity) in the first place? This only shows $Hom(H(C),G) leftrightarrow H^1(C;G) approx H^1(C',G) = coker(i_{n-1}')$?
$endgroup$
– Hawk
Jan 27 at 6:31
|
show 6 more comments
Your Answer
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1 Answer
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1 Answer
1
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active
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$begingroup$
Math: the claim is that if we change the chain complex $C$ and pick a totally different complex C', requiring only that it have the same homology in the $n-1$ spot, then the complex $text{Hom}(C, G)$ and the complex $text{Hom}(C', G)$ would still have the same $text{cok}(iota^*_{n-1})$ (up to canonical isomorphism). The claim is attained by taking the sequence (vi) and the analogous sequence (vi') for $C'$ and applying part two of the lemma.
Philosophy: if this is too much arrow-ception you might try skipping these two pages and just reading the statement of the universal coefficient theorem and doing a couple exercises, then coming back to it. It's one of those theorems that you kind of have to prove for yourself instead of following someone else's proof because it's 8 million steps in a row, any individual one of which isn't too bad, but all of which together are confusing the first 7 to 12 times.
answered Jan 27 at 5:50
hunterhunter
15.3k32640
$endgroup$
$begingroup$
No I know what the lemma is saying. I am asking which specific sequence they are talking about. If you are taking (vi), what is the (vi')? This doesn't exist, it is artificial.
$endgroup$
– Hawk
Jan 27 at 6:09
$begingroup$
(vi') is (vi) but for $C'$ instead of $C$. That is what it means to say that the cokernel doesn't depend on the complex but only the homology.
$endgroup$
– hunter
Jan 27 at 6:18
$begingroup$
So you are telling me we are putting the same sequence by itself and all the vertical maps are identity...?
$endgroup$
– Hawk
Jan 27 at 6:25
$begingroup$
@Hawk I am saying we are taking two different resolutions of $H_{n-1}(C)$. the vertical maps aren't the identity (except the third vertical map, which is).
$endgroup$
– hunter
Jan 27 at 6:27
$begingroup$
$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 $ and this mysterious notational $0 to B_{n-1}' to Z_{n-1}' to H_{n-1}(C') to 0 $ row by row...? How do we know $C to C'$ is an isomorphism (or I guess identity) in the first place? This only shows $Hom(H(C),G) leftrightarrow H^1(C;G) approx H^1(C',G) = coker(i_{n-1}')$?
$endgroup$
– Hawk
Jan 27 at 6:31
|
show 6 more comments
$begingroup$
Math: the claim is that if we change the chain complex $C$ and pick a totally different complex C', requiring only that it have the same homology in the $n-1$ spot, then the complex $text{Hom}(C, G)$ and the complex $text{Hom}(C', G)$ would still have the same $text{cok}(iota^*_{n-1})$ (up to canonical isomorphism). The claim is attained by taking the sequence (vi) and the analogous sequence (vi') for $C'$ and applying part two of the lemma.
Philosophy: if this is too much arrow-ception you might try skipping these two pages and just reading the statement of the universal coefficient theorem and doing a couple exercises, then coming back to it. It's one of those theorems that you kind of have to prove for yourself instead of following someone else's proof because it's 8 million steps in a row, any individual one of which isn't too bad, but all of which together are confusing the first 7 to 12 times.
answered Jan 27 at 5:50
hunterhunter
15.3k32640
$endgroup$
$begingroup$
No I know what the lemma is saying. I am asking which specific sequence they are talking about. If you are taking (vi), what is the (vi')? This doesn't exist, it is artificial.
$endgroup$
– Hawk
Jan 27 at 6:09
$begingroup$
(vi') is (vi) but for $C'$ instead of $C$. That is what it means to say that the cokernel doesn't depend on the complex but only the homology.
$endgroup$
– hunter
Jan 27 at 6:18
$begingroup$
So you are telling me we are putting the same sequence by itself and all the vertical maps are identity...?
$endgroup$
– Hawk
Jan 27 at 6:25
$begingroup$
@Hawk I am saying we are taking two different resolutions of $H_{n-1}(C)$. the vertical maps aren't the identity (except the third vertical map, which is).
$endgroup$
– hunter
Jan 27 at 6:27
$begingroup$
$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 $ and this mysterious notational $0 to B_{n-1}' to Z_{n-1}' to H_{n-1}(C') to 0 $ row by row...? How do we know $C to C'$ is an isomorphism (or I guess identity) in the first place? This only shows $Hom(H(C),G) leftrightarrow H^1(C;G) approx H^1(C',G) = coker(i_{n-1}')$?
$endgroup$
– Hawk
Jan 27 at 6:31
|
show 6 more comments
$begingroup$
Math: the claim is that if we change the chain complex $C$ and pick a totally different complex C', requiring only that it have the same homology in the $n-1$ spot, then the complex $text{Hom}(C, G)$ and the complex $text{Hom}(C', G)$ would still have the same $text{cok}(iota^*_{n-1})$ (up to canonical isomorphism). The claim is attained by taking the sequence (vi) and the analogous sequence (vi') for $C'$ and applying part two of the lemma.
Philosophy: if this is too much arrow-ception you might try skipping these two pages and just reading the statement of the universal coefficient theorem and doing a couple exercises, then coming back to it. It's one of those theorems that you kind of have to prove for yourself instead of following someone else's proof because it's 8 million steps in a row, any individual one of which isn't too bad, but all of which together are confusing the first 7 to 12 times.
answered Jan 27 at 5:50
hunterhunter
15.3k32640
$endgroup$
Math: the claim is that if we change the chain complex $C$ and pick a totally different complex C', requiring only that it have the same homology in the $n-1$ spot, then the complex $text{Hom}(C, G)$ and the complex $text{Hom}(C', G)$ would still have the same $text{cok}(iota^*_{n-1})$ (up to canonical isomorphism). The claim is attained by taking the sequence (vi) and the analogous sequence (vi') for $C'$ and applying part two of the lemma.
Philosophy: if this is too much arrow-ception you might try skipping these two pages and just reading the statement of the universal coefficient theorem and doing a couple exercises, then coming back to it. It's one of those theorems that you kind of have to prove for yourself instead of following someone else's proof because it's 8 million steps in a row, any individual one of which isn't too bad, but all of which together are confusing the first 7 to 12 times.
answered Jan 27 at 5:50
hunterhunter
15.3k32640
answered Jan 27 at 5:50
hunterhunter
15.3k32640
answered Jan 27 at 5:50
hunterhunter
15.3k32640
answered Jan 27 at 5:50
hunterhunter
15.3k32640
15.3k32640
$begingroup$
No I know what the lemma is saying. I am asking which specific sequence they are talking about. If you are taking (vi), what is the (vi')? This doesn't exist, it is artificial.
$endgroup$
– Hawk
Jan 27 at 6:09
$begingroup$
(vi') is (vi) but for $C'$ instead of $C$. That is what it means to say that the cokernel doesn't depend on the complex but only the homology.
$endgroup$
– hunter
Jan 27 at 6:18
$begingroup$
So you are telling me we are putting the same sequence by itself and all the vertical maps are identity...?
$endgroup$
– Hawk
Jan 27 at 6:25
$begingroup$
@Hawk I am saying we are taking two different resolutions of $H_{n-1}(C)$. the vertical maps aren't the identity (except the third vertical map, which is).
$endgroup$
– hunter
Jan 27 at 6:27
$begingroup$
$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 $ and this mysterious notational $0 to B_{n-1}' to Z_{n-1}' to H_{n-1}(C') to 0 $ row by row...? How do we know $C to C'$ is an isomorphism (or I guess identity) in the first place? This only shows $Hom(H(C),G) leftrightarrow H^1(C;G) approx H^1(C',G) = coker(i_{n-1}')$?
$endgroup$
– Hawk
Jan 27 at 6:31
|
show 6 more comments
$begingroup$
No I know what the lemma is saying. I am asking which specific sequence they are talking about. If you are taking (vi), what is the (vi')? This doesn't exist, it is artificial.
$endgroup$
– Hawk
Jan 27 at 6:09
$begingroup$
(vi') is (vi) but for $C'$ instead of $C$. That is what it means to say that the cokernel doesn't depend on the complex but only the homology.
$endgroup$
– hunter
Jan 27 at 6:18
$begingroup$
So you are telling me we are putting the same sequence by itself and all the vertical maps are identity...?
$endgroup$
– Hawk
Jan 27 at 6:25
$begingroup$
@Hawk I am saying we are taking two different resolutions of $H_{n-1}(C)$. the vertical maps aren't the identity (except the third vertical map, which is).
$endgroup$
– hunter
Jan 27 at 6:27
$begingroup$
$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 $ and this mysterious notational $0 to B_{n-1}' to Z_{n-1}' to H_{n-1}(C') to 0 $ row by row...? How do we know $C to C'$ is an isomorphism (or I guess identity) in the first place? This only shows $Hom(H(C),G) leftrightarrow H^1(C;G) approx H^1(C',G) = coker(i_{n-1}')$?
$endgroup$
– Hawk
Jan 27 at 6:31
$begingroup$
No I know what the lemma is saying. I am asking which specific sequence they are talking about. If you are taking (vi), what is the (vi')? This doesn't exist, it is artificial.
$endgroup$
– Hawk
Jan 27 at 6:09
$begingroup$
No I know what the lemma is saying. I am asking which specific sequence they are talking about. If you are taking (vi), what is the (vi')? This doesn't exist, it is artificial.
$endgroup$
– Hawk
Jan 27 at 6:09
$begingroup$
(vi') is (vi) but for $C'$ instead of $C$. That is what it means to say that the cokernel doesn't depend on the complex but only the homology.
$endgroup$
– hunter
Jan 27 at 6:18
$begingroup$
(vi') is (vi) but for $C'$ instead of $C$. That is what it means to say that the cokernel doesn't depend on the complex but only the homology.
$endgroup$
– hunter
Jan 27 at 6:18
$begingroup$
So you are telling me we are putting the same sequence by itself and all the vertical maps are identity...?
$endgroup$
– Hawk
Jan 27 at 6:25
$begingroup$
So you are telling me we are putting the same sequence by itself and all the vertical maps are identity...?
$endgroup$
– Hawk
Jan 27 at 6:25
$begingroup$
@Hawk I am saying we are taking two different resolutions of $H_{n-1}(C)$. the vertical maps aren't the identity (except the third vertical map, which is).
$endgroup$
– hunter
Jan 27 at 6:27
$begingroup$
@Hawk I am saying we are taking two different resolutions of $H_{n-1}(C)$. the vertical maps aren't the identity (except the third vertical map, which is).
$endgroup$
– hunter
Jan 27 at 6:27
$begingroup$
$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 $ and this mysterious notational $0 to B_{n-1}' to Z_{n-1}' to H_{n-1}(C') to 0 $ row by row...? How do we know $C to C'$ is an isomorphism (or I guess identity) in the first place? This only shows $Hom(H(C),G) leftrightarrow H^1(C;G) approx H^1(C',G) = coker(i_{n-1}')$?
$endgroup$
– Hawk
Jan 27 at 6:31
$begingroup$
$0 to B_{n-1} to Z_{n-1} to H_{n-1}(C) to 0 $ and this mysterious notational $0 to B_{n-1}' to Z_{n-1}' to H_{n-1}(C') to 0 $ row by row...? How do we know $C to C'$ is an isomorphism (or I guess identity) in the first place? This only shows $Hom(H(C),G) leftrightarrow H^1(C;G) approx H^1(C',G) = coker(i_{n-1}')$?
$endgroup$
– Hawk
Jan 27 at 6:31
|
show 6 more comments
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