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Tom Goodwillie
Tom Goodwillie
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No. Let $R$ be commutative, $T=R\times R$, $s(x,y)=(y,x)$. Then $T\otimes T^{op}=T\otimes T$ is the product of four copies of $R\otimes R$, so that $K(T\otimes T^{op})$ is the product of four copies of $K(R\otimes R)$ and $h$ simply permutes the copies in some nontrivial way.

No. Let $R$ be commutative, $T=R\times R$, $s(x,y)=(y,x)$. Then $T\otimes T^{op}=T\otimes T$ is the product of four copies of $R\otimes R$, so that $K(T\otimes T^{op})$ is the product of four copies of $K(R\otimes R)$ and $h$ simply permutes the copies in some nontrivial way.

No. Let $R$ be commutative, $T=R\times R$, $s(x,y)=(y,x)$. Then $T\otimes T^{op}=T\otimes T$ is the product of four copies of $R\otimes R$, so that $K(T\otimes T^{op})$ is the product of four copies of $K(R\otimes R)$ and $h$ permutes the copies in some nontrivial way.

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Tom Goodwillie
Tom Goodwillie
  • 55.9k
  • 7
  • 151
  • 240

No. Let $R$ be commutative, $T=R\times R$, $s(x,y)=(y,x)$. Then $T\otimes T^{op}=T\otimes T$ is the product of four copies of $R\otimes R$, so that $K(T\otimes T^{op})$ is the product of four copies of $K(R\otimes R)$ and $h$ simply permutes the copies in some nontrivial way.