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.



Involution map, and induced morphism in K-theory
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.


