Involution map, and induced morphism in K-theory

Question

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.

Answer 1

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.

Answer 2

Let $T = \mathbb C[\mathbb Z/3]$, and $s$ be the involution induced by the nontrivial isomorphism. Then $T \otimes T^{\rm op}$ is $\mathbb C [ \mathbb Z/3 \times \mathbb Z/3]$ and $K_0(T \otimes T^{\rm op})$ is the ring $\mathbb Z[\mu_3 \times \mu_3]$. Then $h$ acts on this ring by the homomorphism induced by $(\zeta_1, \zeta_2) \mapsto (\overline \zeta_2, \overline \zeta_1)$. This homomorphism is not trivial.