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user163897
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(This is based on Heller's simple proof, according to Rotman)

$A, B$ are a complex and its subcomplex of boundaries. Assume all the modules in $A, B$ are flat.

We have an exact sequence of complexes: $0\to Z \to A \to B[-1] \to 0$ and after tensoring $C$ where $C$ is any complex. $0\to Z\otimes C \to A\otimes C \to B[-1] \otimes C \to 0$.

The corresponding long exact sequence is:

$H_{n+1}(B[-1]\otimes C) \to H_n(Z\otimes C) \to H_n(A\otimes C)\to H_{n}(B[-1]\otimes C) \to ...$

Then $H_{n+1}(B[-1]\otimes C) = (B \otimes C)_n$ and $H_n(Z \otimes C) = (Z \otimes C)_n$ since $Z, B[-1]$ have zero differentials. Thus, we may rewrite the long exact sequence as:

$(B \otimes C)_n \to (Z \otimes C)_n \to H_n(A\otimes C)\to (B \otimes C)_{n-1} \to ...$

I got a bit confused. I think as a total complex, the differential of $Z \otimes C$ should be of the form $a_{p}\otimes b_{q} \mapsto \Delta'a_{p}\otimes b_{q} + (-1)^p a_{p}\otimes \Delta''_{q}b_{q}$. So $\Delta'=0$ but we still have $\Delta''$ here, so it should not be a zero-map...

Any help would be appreicated!

$A, B$ are a complex and its subcomplex of boundaries. Assume all the modules in $A, B$ are flat.

We have an exact sequence of complexes: $0\to Z \to A \to B[-1] \to 0$ and after tensoring $C$ where $C$ is any complex. $0\to Z\otimes C \to A\otimes C \to B[-1] \otimes C \to 0$.

The corresponding long exact sequence is:

$H_{n+1}(B[-1]\otimes C) \to H_n(Z\otimes C) \to H_n(A\otimes C)\to H_{n}(B[-1]\otimes C) \to ...$

Then $H_{n+1}(B[-1]\otimes C) = (B \otimes C)_n$ and $H_n(Z \otimes C) = (Z \otimes C)_n$ since $Z, B[-1]$ have zero differentials.

I got a bit confused. I think as a total complex, the differential of $Z \otimes C$ should be of the form $a_{p}\otimes b_{q} \mapsto \Delta'a_{p}\otimes b_{q} + (-1)^p a_{p}\otimes \Delta''_{q}b_{q}$. So $\Delta'=0$ but we still have $\Delta''$ here, so it should not be a zero-map...

Any help would be appreicated!

(This is based on Heller's simple proof, according to Rotman)

$A, B$ are a complex and its subcomplex of boundaries. Assume all the modules in $A, B$ are flat.

We have an exact sequence of complexes: $0\to Z \to A \to B[-1] \to 0$ and after tensoring $C$ where $C$ is any complex. $0\to Z\otimes C \to A\otimes C \to B[-1] \otimes C \to 0$.

The corresponding long exact sequence is:

$H_{n+1}(B[-1]\otimes C) \to H_n(Z\otimes C) \to H_n(A\otimes C)\to H_{n}(B[-1]\otimes C) \to ...$

Then $H_{n+1}(B[-1]\otimes C) = (B \otimes C)_n$ and $H_n(Z \otimes C) = (Z \otimes C)_n$ since $Z, B[-1]$ have zero differentials. Thus, we may rewrite the long exact sequence as:

$(B \otimes C)_n \to (Z \otimes C)_n \to H_n(A\otimes C)\to (B \otimes C)_{n-1} \to ...$

I got a bit confused. I think as a total complex, the differential of $Z \otimes C$ should be of the form $a_{p}\otimes b_{q} \mapsto \Delta'a_{p}\otimes b_{q} + (-1)^p a_{p}\otimes \Delta''_{q}b_{q}$. So $\Delta'=0$ but we still have $\Delta''$ here, so it should not be a zero-map...

Any help would be appreicated!

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user163897
user163897
  • 41
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$A, Z$$A, B$ are a complex and its cycle subcomplex of boundaries. Assume all the modules in $R$-modules$A, B$ are flat. We

We have an exact sequence of complexes: $0\to Z\otimes C \to A\otimes C \to B[-1] \otimes C \to 0$$0\to Z \to A \to B[-1] \to 0$ and after tensoring $C$ where $C$ is any complex. Then$0\to Z\otimes C \to A\otimes C \to B[-1] \otimes C \to 0$.

The corresponding long exact sequence is:

$H_{n+1}(B[-1]\otimes C) \to H_n(Z\otimes C) \to H_n(A\otimes C)\to H_{n}(B[-1]\otimes C) \to ...$

Then $H_{n+1}(B[-1]\otimes C) = (B \otimes C)_n$ and $H_n(Z \otimes C) = (Z \otimes C)_n$ since $Z$ has$Z, B[-1]$ have zero differentials.

I got a bit confused. I think as a total complex, the differential of $Z \otimes C$ should be of the form $a_{p}\otimes b_{q} \mapsto \Delta'a_{p}\otimes b_{q} + (-1)^p a_{p}\otimes \Delta''_{q}b_{q}$. So $\Delta'=0$ but we still have $\Delta''$ here, so it should not be a zero-map... Any

Any help would be appreicated!

$A, Z$ are a complex and its cycle subcomplex of $R$-modules. We have an exact sequence of complexes: $0\to Z\otimes C \to A\otimes C \to B[-1] \otimes C \to 0$ where $C$ is any complex. Then $H_n(Z \otimes C) = (Z \otimes C)_n$ since $Z$ has zero differentials.

I got a bit confused. I think as a total complex, the differential of $Z \otimes C$ should be of the form $a_{p}\otimes b_{q} \mapsto \Delta'a_{p}\otimes b_{q} + (-1)^p a_{p}\otimes \Delta''_{q}b_{q}$. So $\Delta'=0$ but we still have $\Delta''$ here, so it should not be a zero-map... Any help would be appreicated!

$A, B$ are a complex and its subcomplex of boundaries. Assume all the modules in $A, B$ are flat.

We have an exact sequence of complexes: $0\to Z \to A \to B[-1] \to 0$ and after tensoring $C$ where $C$ is any complex. $0\to Z\otimes C \to A\otimes C \to B[-1] \otimes C \to 0$.

The corresponding long exact sequence is:

$H_{n+1}(B[-1]\otimes C) \to H_n(Z\otimes C) \to H_n(A\otimes C)\to H_{n}(B[-1]\otimes C) \to ...$

Then $H_{n+1}(B[-1]\otimes C) = (B \otimes C)_n$ and $H_n(Z \otimes C) = (Z \otimes C)_n$ since $Z, B[-1]$ have zero differentials.

I got a bit confused. I think as a total complex, the differential of $Z \otimes C$ should be of the form $a_{p}\otimes b_{q} \mapsto \Delta'a_{p}\otimes b_{q} + (-1)^p a_{p}\otimes \Delta''_{q}b_{q}$. So $\Delta'=0$ but we still have $\Delta''$ here, so it should not be a zero-map...

Any help would be appreicated!

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user163897
user163897
  • 41
  • 2

$H_n(Z \otimes C) = (Z \otimes C)_n$

$A, Z$ are a complex and its cycle subcomplex of $R$-modules. We have an exact sequence of complexes: $0\to Z\otimes C \to A\otimes C \to B[-1] \otimes C \to 0$ where $C$ is any complex. Then $H_n(Z \otimes C) = (Z \otimes C)_n$ since $Z$ has zero differentials.

I got a bit confused. I think as a total complex, the differential of $Z \otimes C$ should be of the form $a_{p}\otimes b_{q} \mapsto \Delta'a_{p}\otimes b_{q} + (-1)^p a_{p}\otimes \Delta''_{q}b_{q}$. So $\Delta'=0$ but we still have $\Delta''$ here, so it should not be a zero-map... Any help would be appreicated!