Skip to main content
formatting, added tag
Source Link
YCor
  • 63.9k
  • 5
  • 187
  • 285

Kunneth Künneth formula for cohomology

Is$\DeclareMathOperator\Hom{Hom}$Is there an algebraic KunnethKünneth formula for cohomology?

More precisely assume $A_{*}, B_{*}$ are chain complexes of free $R$-modules ($R$ is a $PID$) and $M, N$ are $R$-modules. Then the map $\sum H^n(A_{*},M)\otimes H^m(B_{*},N)\rightarrow H^{n+m}(A_{*}\otimes B_{*}, M\otimes N)$ is defined as usually.

Is there an exact sequence of $R$-modules involving the map above analogous to the corresponding well-known KunnethKünneth formulas for homology and universal coefficients theorems for homology and cohomology?

The problem here which confuses me is that in general for two free $R$-modules $A$ and $B$, $Hom(A,M)\otimes Hom(B,N)\neq Hom(A\otimes B, M\otimes N)$$\Hom(A,M)\otimes \Hom(B,N)\neq \Hom(A\otimes B, M\otimes N)$ so one can not just take the cochain complexes $Hom(A_{*}, M), Hom(B_{*}, N)$$\Hom(A_{*}, M), \Hom(B_{*}, N)$ then consider them as chain complexes with the "reversed" order and apply a usual KunnethKünneth formula for homology as was suggested for example in J.P.May "Coincise May "Concise course of algebraic topology".

This strategy would work say for cellular cohomology of finite $CW$CW-complexes but not in general.

Kunneth formula for cohomology

Is there an algebraic Kunneth formula for cohomology?

More precisely assume $A_{*}, B_{*}$ are chain complexes of free $R$-modules ($R$ is a $PID$) and $M, N$ are $R$-modules. Then the map $\sum H^n(A_{*},M)\otimes H^m(B_{*},N)\rightarrow H^{n+m}(A_{*}\otimes B_{*}, M\otimes N)$ is defined as usually.

Is there an exact sequence of $R$-modules involving the map above analogous to the corresponding well-known Kunneth formulas for homology and universal coefficients theorems for homology and cohomology?

The problem here which confuses me is that in general for two free $R$-modules $A$ and $B$, $Hom(A,M)\otimes Hom(B,N)\neq Hom(A\otimes B, M\otimes N)$ so one can not just take the cochain complexes $Hom(A_{*}, M), Hom(B_{*}, N)$ then consider them as chain complexes with the "reversed" order and apply a usual Kunneth formula for homology as was suggested for example in J.P.May "Coincise course of algebraic topology".

This strategy would work say for cellular cohomology of finite $CW$-complexes but not in general.

Künneth formula for cohomology

$\DeclareMathOperator\Hom{Hom}$Is there an algebraic Künneth formula for cohomology?

More precisely assume $A_{*}, B_{*}$ are chain complexes of free $R$-modules ($R$ is a $PID$) and $M, N$ are $R$-modules. Then the map $\sum H^n(A_{*},M)\otimes H^m(B_{*},N)\rightarrow H^{n+m}(A_{*}\otimes B_{*}, M\otimes N)$ is defined as usually.

Is there an exact sequence of $R$-modules involving the map above analogous to the corresponding well-known Künneth formulas for homology and universal coefficients theorems for homology and cohomology?

The problem here which confuses me is that in general for two free $R$-modules $A$ and $B$, $\Hom(A,M)\otimes \Hom(B,N)\neq \Hom(A\otimes B, M\otimes N)$ so one can not just take the cochain complexes $\Hom(A_{*}, M), \Hom(B_{*}, N)$ then consider them as chain complexes with the "reversed" order and apply a usual Künneth formula for homology as was suggested for example in J.P. May "Concise course of algebraic topology".

This strategy would work say for cellular cohomology of finite CW-complexes but not in general.

Source Link
Axel
  • 1.4k
  • 2
  • 12
  • 14

Kunneth formula for cohomology

Is there an algebraic Kunneth formula for cohomology?

More precisely assume $A_{*}, B_{*}$ are chain complexes of free $R$-modules ($R$ is a $PID$) and $M, N$ are $R$-modules. Then the map $\sum H^n(A_{*},M)\otimes H^m(B_{*},N)\rightarrow H^{n+m}(A_{*}\otimes B_{*}, M\otimes N)$ is defined as usually.

Is there an exact sequence of $R$-modules involving the map above analogous to the corresponding well-known Kunneth formulas for homology and universal coefficients theorems for homology and cohomology?

The problem here which confuses me is that in general for two free $R$-modules $A$ and $B$, $Hom(A,M)\otimes Hom(B,N)\neq Hom(A\otimes B, M\otimes N)$ so one can not just take the cochain complexes $Hom(A_{*}, M), Hom(B_{*}, N)$ then consider them as chain complexes with the "reversed" order and apply a usual Kunneth formula for homology as was suggested for example in J.P.May "Coincise course of algebraic topology".

This strategy would work say for cellular cohomology of finite $CW$-complexes but not in general.