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Proofreading; `^{op}` -> `^\text{op}`
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LSpice
LSpice
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Involution map, and induced morphism in kK-theory

Let $T$ be a ring with involution $s:T\rightarrow T$. And let $$h:T\otimes T^{op} \rightarrow T\otimes T^{op}$$$$h:T\otimes T^\text{op} \rightarrow T\otimes T^\text{op}$$ be the automorphism of ring automorphism given by $h(a\otimes b)=s(b)\otimes s(a)$.

I was wanderingwondering if the induced homorphism $$ K_{\ast}(h): K_{\ast}(T\otimes T^{op})\rightarrow K_{\ast}(T\otimes T^{op}) $$$$ K_{\ast}(h): K_{\ast}(T\otimes T^\text{op})\rightarrow K_{\ast}(T\otimes T^\text{op}) $$ in K-theory, is the identity map?

Notice that "op" is used for the opposite multiplication.

Involution map, and induced morphism in k-theory

Let $T$ be a ring with involution $s:T\rightarrow T$. And let $$h:T\otimes T^{op} \rightarrow T\otimes T^{op}$$ the automorphism of ring given by $h(a\otimes b)=s(b)\otimes s(a)$.

I was wandering if the induced homorphism $$ K_{\ast}(h): K_{\ast}(T\otimes T^{op})\rightarrow K_{\ast}(T\otimes T^{op}) $$ in K-theory, is the identity map?

Notice that "op" is used for the opposite multiplication.

Involution map, and induced morphism in K-theory

Let $T$ be a ring with involution $s:T\rightarrow T$. And let $$h:T\otimes T^\text{op} \rightarrow T\otimes T^\text{op}$$ be the ring automorphism given by $h(a\otimes b)=s(b)\otimes s(a)$.

I was wondering if the induced homorphism $$ K_{\ast}(h): K_{\ast}(T\otimes T^\text{op})\rightarrow K_{\ast}(T\otimes T^\text{op}) $$ in K-theory, is the identity map?

Notice that "op" is used for the opposite multiplication.

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LGO
LGO
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Involution map, and induced morphism in k-theory

Let $T$ be a ring with involution $s:T\rightarrow T$. And let $$h:T\otimes T^{op} \rightarrow T\otimes T^{op}$$ the automorphism of ring given by $h(a\otimes b)=s(b)\otimes s(a)$.

I was wandering if the induced homorphism $$ K_{\ast}(h): K_{\ast}(T\otimes T^{op})\rightarrow K_{\ast}(T\otimes T^{op}) $$ in K-theory, is the identity map?

Notice that "op" is used for the opposite multiplication.