Mixing
How to get the cohomology group in the exact sequence? $B_n^* leftarrow Z_n^* leftarrow H^n(C;G) leftarrow...
$begingroup$
In Hatcher page 192, the sequence is extracted
$B_n^* leftarrow Z_n^* leftarrow H^n(C;G) leftarrow B_{n-1}^* leftarrow Z_{n-1}^*$ from the diagram
$Z_n^* = ker delta_n$, $B_n^* = Img delta_{n-1}$ are the dual spaces of the spaces without $*$
Let explain as I myself found it terse. This dual map is obtained from the split exactness of the row. This is split because $B_n$ is free. So the "pulling" (not pullback) described is from the split exactness. I am not quite sure what he meant by undoing here, but in any case that is the construction.
Basically my question is, how do we get the $H^n(C;G)$ in the middle of the sequence $B_n^* leftarrow Z_n^* leftarrow H^n(C;G) leftarrow B_{n-1}^* leftarrow Z_{n-1}^*$?
algebraic-topology proof-explanation homology-cohomology
asked Jan 20 at 6:46
HawkHawk
5,5401139107
$endgroup$
add a comment |
$begingroup$
In Hatcher page 192, the sequence is extracted
$B_n^* leftarrow Z_n^* leftarrow H^n(C;G) leftarrow B_{n-1}^* leftarrow Z_{n-1}^*$ from the diagram
$Z_n^* = ker delta_n$, $B_n^* = Img delta_{n-1}$ are the dual spaces of the spaces without $*$
Let explain as I myself found it terse. This dual map is obtained from the split exactness of the row. This is split because $B_n$ is free. So the "pulling" (not pullback) described is from the split exactness. I am not quite sure what he meant by undoing here, but in any case that is the construction.
Basically my question is, how do we get the $H^n(C;G)$ in the middle of the sequence $B_n^* leftarrow Z_n^* leftarrow H^n(C;G) leftarrow B_{n-1}^* leftarrow Z_{n-1}^*$?
algebraic-topology proof-explanation homology-cohomology
asked Jan 20 at 6:46
HawkHawk
5,5401139107
$endgroup$
add a comment |
$begingroup$
In Hatcher page 192, the sequence is extracted
$B_n^* leftarrow Z_n^* leftarrow H^n(C;G) leftarrow B_{n-1}^* leftarrow Z_{n-1}^*$ from the diagram
$Z_n^* = ker delta_n$, $B_n^* = Img delta_{n-1}$ are the dual spaces of the spaces without $*$
Let explain as I myself found it terse. This dual map is obtained from the split exactness of the row. This is split because $B_n$ is free. So the "pulling" (not pullback) described is from the split exactness. I am not quite sure what he meant by undoing here, but in any case that is the construction.
Basically my question is, how do we get the $H^n(C;G)$ in the middle of the sequence $B_n^* leftarrow Z_n^* leftarrow H^n(C;G) leftarrow B_{n-1}^* leftarrow Z_{n-1}^*$?
algebraic-topology proof-explanation homology-cohomology
asked Jan 20 at 6:46
HawkHawk
5,5401139107
$endgroup$
In Hatcher page 192, the sequence is extracted
$B_n^* leftarrow Z_n^* leftarrow H^n(C;G) leftarrow B_{n-1}^* leftarrow Z_{n-1}^*$ from the diagram
$Z_n^* = ker delta_n$, $B_n^* = Img delta_{n-1}$ are the dual spaces of the spaces without $*$
Let explain as I myself found it terse. This dual map is obtained from the split exactness of the row. This is split because $B_n$ is free. So the "pulling" (not pullback) described is from the split exactness. I am not quite sure what he meant by undoing here, but in any case that is the construction.
Basically my question is, how do we get the $H^n(C;G)$ in the middle of the sequence $B_n^* leftarrow Z_n^* leftarrow H^n(C;G) leftarrow B_{n-1}^* leftarrow Z_{n-1}^*$?
algebraic-topology proof-explanation homology-cohomology
algebraic-topology proof-explanation homology-cohomology
asked Jan 20 at 6:46
HawkHawk
5,5401139107
asked Jan 20 at 6:46
HawkHawk
5,5401139107
asked Jan 20 at 6:46
HawkHawk
5,5401139107
asked Jan 20 at 6:46
HawkHawk
5,5401139107
asked Jan 20 at 6:46
HawkHawk
5,5401139107
5,5401139107
add a comment |
add a comment |
1 Answer
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$begingroup$
As far as I can tell, this is just part of the long exact sequence of homology
associated to the diagram (ii). In (ii) the homology of the vertical column
at $C_n^*$ is $H^n(C;G)$ (I presume Hatcher defined $H^n(C;G)$ thusly).
The homology of the left column is just the $Z_n^*$, since the differentials
are zero. Similarly for the right column.
answered Jan 20 at 6:56
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
No, we define $C_n^* = Hom(C_n,G)$, but I think what he did show was that there is a homomorphism $H(C;G) to Hom(H_n(C),G)$ and vice versa.
$endgroup$
– Hawk
Jan 20 at 7:40
$begingroup$
Nvm I figured it out, and actually I just read this answer again and realize that's what you are telling me. For some reason when the sequence is presented to me horizontally, my brain just doesn't recognize the associated long sequence.
$endgroup$
– Hawk
Jan 23 at 4:32
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
As far as I can tell, this is just part of the long exact sequence of homology
associated to the diagram (ii). In (ii) the homology of the vertical column
at $C_n^*$ is $H^n(C;G)$ (I presume Hatcher defined $H^n(C;G)$ thusly).
The homology of the left column is just the $Z_n^*$, since the differentials
are zero. Similarly for the right column.
answered Jan 20 at 6:56
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
No, we define $C_n^* = Hom(C_n,G)$, but I think what he did show was that there is a homomorphism $H(C;G) to Hom(H_n(C),G)$ and vice versa.
$endgroup$
– Hawk
Jan 20 at 7:40
$begingroup$
Nvm I figured it out, and actually I just read this answer again and realize that's what you are telling me. For some reason when the sequence is presented to me horizontally, my brain just doesn't recognize the associated long sequence.
$endgroup$
– Hawk
Jan 23 at 4:32
add a comment |
$begingroup$
As far as I can tell, this is just part of the long exact sequence of homology
associated to the diagram (ii). In (ii) the homology of the vertical column
at $C_n^*$ is $H^n(C;G)$ (I presume Hatcher defined $H^n(C;G)$ thusly).
The homology of the left column is just the $Z_n^*$, since the differentials
are zero. Similarly for the right column.
answered Jan 20 at 6:56
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
No, we define $C_n^* = Hom(C_n,G)$, but I think what he did show was that there is a homomorphism $H(C;G) to Hom(H_n(C),G)$ and vice versa.
$endgroup$
– Hawk
Jan 20 at 7:40
$begingroup$
Nvm I figured it out, and actually I just read this answer again and realize that's what you are telling me. For some reason when the sequence is presented to me horizontally, my brain just doesn't recognize the associated long sequence.
$endgroup$
– Hawk
Jan 23 at 4:32
add a comment |
$begingroup$
As far as I can tell, this is just part of the long exact sequence of homology
associated to the diagram (ii). In (ii) the homology of the vertical column
at $C_n^*$ is $H^n(C;G)$ (I presume Hatcher defined $H^n(C;G)$ thusly).
The homology of the left column is just the $Z_n^*$, since the differentials
are zero. Similarly for the right column.
answered Jan 20 at 6:56
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
As far as I can tell, this is just part of the long exact sequence of homology
associated to the diagram (ii). In (ii) the homology of the vertical column
at $C_n^*$ is $H^n(C;G)$ (I presume Hatcher defined $H^n(C;G)$ thusly).
The homology of the left column is just the $Z_n^*$, since the differentials
are zero. Similarly for the right column.
answered Jan 20 at 6:56
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 20 at 6:56
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 20 at 6:56
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 20 at 6:56
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
No, we define $C_n^* = Hom(C_n,G)$, but I think what he did show was that there is a homomorphism $H(C;G) to Hom(H_n(C),G)$ and vice versa.
$endgroup$
– Hawk
Jan 20 at 7:40
$begingroup$
Nvm I figured it out, and actually I just read this answer again and realize that's what you are telling me. For some reason when the sequence is presented to me horizontally, my brain just doesn't recognize the associated long sequence.
$endgroup$
– Hawk
Jan 23 at 4:32
add a comment |
$begingroup$
No, we define $C_n^* = Hom(C_n,G)$, but I think what he did show was that there is a homomorphism $H(C;G) to Hom(H_n(C),G)$ and vice versa.
$endgroup$
– Hawk
Jan 20 at 7:40
$begingroup$
Nvm I figured it out, and actually I just read this answer again and realize that's what you are telling me. For some reason when the sequence is presented to me horizontally, my brain just doesn't recognize the associated long sequence.
$endgroup$
– Hawk
Jan 23 at 4:32
$begingroup$
No, we define $C_n^* = Hom(C_n,G)$, but I think what he did show was that there is a homomorphism $H(C;G) to Hom(H_n(C),G)$ and vice versa.
$endgroup$
– Hawk
Jan 20 at 7:40
$begingroup$
No, we define $C_n^* = Hom(C_n,G)$, but I think what he did show was that there is a homomorphism $H(C;G) to Hom(H_n(C),G)$ and vice versa.
$endgroup$
– Hawk
Jan 20 at 7:40
$begingroup$
Nvm I figured it out, and actually I just read this answer again and realize that's what you are telling me. For some reason when the sequence is presented to me horizontally, my brain just doesn't recognize the associated long sequence.
$endgroup$
– Hawk
Jan 23 at 4:32
$begingroup$
Nvm I figured it out, and actually I just read this answer again and realize that's what you are telling me. For some reason when the sequence is presented to me horizontally, my brain just doesn't recognize the associated long sequence.
$endgroup$
– Hawk
Jan 23 at 4:32
add a comment |
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