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 $h(a\otimes b)=s(b)\otimes s(a)$.


**suppose that $$ K\_{0}(h): K\_{0}(T\otimes T^\text{op})\rightarrow K\_{0}(T\otimes T^\text{op}) $$ is the identity map.**


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.



Induced map in k-theory by an involution
I don't think so: Let us take $T=\mathbb{Z}[x^{\pm 1}]$, and let us take $s$ the identity. Then $T\otimes T^{op}= \mathbb{Z}[x^{\pm 1},y^{\pm 1}]$ has $K\_0 = \mathbb{Z}$ detected by rank (by Grothendieck-Serre), so $K\_0(h)$ is the identity. But I think in $K\_1$ the elements given by the units $x$ and $y$ are different, and are interchanged by $K\_1(h)$.


