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Mark Grant
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Edit: The answer below assumes that $A$ and $B$ are bounded below (or above). This is not explicitly mentioned in the statement in Spanier, but seems to be necessary.


Yes, this is Theorem 5.5.11 in Spanier's "Algebraic Topology" text.

The conditions are that the torsion product $\operatorname{Tor}_R(M,N)=0$ and either $H(A;M)$ and $H(B;N)$ are of finite type, or $H(B;N)$ is of finite type and $N$ is finitely generated.

Then there is a natural short exact sequence $$ 0 \to H(A;M)\otimes H(B;N)\to H(A\otimes B; M\otimes N)\to Tor_R(H(A;M),H(B;N))\to 0 $$ (where the second map raises degree by one) and this sequence splits.

Yes, this is Theorem 5.5.11 in Spanier's "Algebraic Topology" text.

The conditions are that the torsion product $\operatorname{Tor}_R(M,N)=0$ and either $H(A;M)$ and $H(B;N)$ are of finite type, or $H(B;N)$ is of finite type and $N$ is finitely generated.

Then there is a natural short exact sequence $$ 0 \to H(A;M)\otimes H(B;N)\to H(A\otimes B; M\otimes N)\to Tor_R(H(A;M),H(B;N))\to 0 $$ (where the second map raises degree by one) and this sequence splits.

Edit: The answer below assumes that $A$ and $B$ are bounded below (or above). This is not explicitly mentioned in the statement in Spanier, but seems to be necessary.


Yes, this is Theorem 5.5.11 in Spanier's "Algebraic Topology" text.

The conditions are that the torsion product $\operatorname{Tor}_R(M,N)=0$ and either $H(A;M)$ and $H(B;N)$ are of finite type, or $H(B;N)$ is of finite type and $N$ is finitely generated.

Then there is a natural short exact sequence $$ 0 \to H(A;M)\otimes H(B;N)\to H(A\otimes B; M\otimes N)\to Tor_R(H(A;M),H(B;N))\to 0 $$ (where the second map raises degree by one) and this sequence splits.

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Mark Grant
  • 35.9k
  • 8
  • 95
  • 198

Yes, this is Theorem 5.5.11 in Spanier's "Algebraic Topology" text.

The conditions are that the torsion product $\operatorname{Tor}_R(M,N)=0$ and either $H(A;M)$ and $H(B;N)$ are of finite type, or $H(B;N)$ is of finite type and $N$ is finitely generated.

Then there is a natural short exact sequence $$ 0 \to H(A;M)\otimes H(B;N)\to H(A\otimes B; M\otimes N)\to Tor_R(H^\ast(A;M),H^\ast(B;N))\to 0 $$$$ 0 \to H(A;M)\otimes H(B;N)\to H(A\otimes B; M\otimes N)\to Tor_R(H(A;M),H(B;N))\to 0 $$ (where the second map raises degree by one) and this sequence splits.

Yes, this is Theorem 5.5.11 in Spanier's "Algebraic Topology" text.

The conditions are that the torsion product $\operatorname{Tor}_R(M,N)=0$ and either $H(A;M)$ and $H(B;N)$ are of finite type, or $H(B;N)$ is of finite type and $N$ is finitely generated.

Then there is a natural short exact sequence $$ 0 \to H(A;M)\otimes H(B;N)\to H(A\otimes B; M\otimes N)\to Tor_R(H^\ast(A;M),H^\ast(B;N))\to 0 $$ (where the second map raises degree by one) and this sequence splits.

Yes, this is Theorem 5.5.11 in Spanier's "Algebraic Topology" text.

The conditions are that the torsion product $\operatorname{Tor}_R(M,N)=0$ and either $H(A;M)$ and $H(B;N)$ are of finite type, or $H(B;N)$ is of finite type and $N$ is finitely generated.

Then there is a natural short exact sequence $$ 0 \to H(A;M)\otimes H(B;N)\to H(A\otimes B; M\otimes N)\to Tor_R(H(A;M),H(B;N))\to 0 $$ (where the second map raises degree by one) and this sequence splits.

Source Link
Mark Grant
  • 35.9k
  • 8
  • 95
  • 198

Yes, this is Theorem 5.5.11 in Spanier's "Algebraic Topology" text.

The conditions are that the torsion product $\operatorname{Tor}_R(M,N)=0$ and either $H(A;M)$ and $H(B;N)$ are of finite type, or $H(B;N)$ is of finite type and $N$ is finitely generated.

Then there is a natural short exact sequence $$ 0 \to H(A;M)\otimes H(B;N)\to H(A\otimes B; M\otimes N)\to Tor_R(H^\ast(A;M),H^\ast(B;N))\to 0 $$ (where the second map raises degree by one) and this sequence splits.