Hello I have the next doubt:
Let $A$ be a commutative ring and $T:=T(A)=\left\{\left( \begin{array}{cc} u & 0 \\ 0 & u^{-1} \\ \end{array} \right) \;:\;u\in A^{\ast}\right\}$ the subgroup of diagonable matrices of $SL_{2}(A)$ where $A^{\ast}$ is the group of units.
Now let $w=\left( \begin{array}{cc} 0&-1 \\ 1&0 \\ \end{array} \right) $ in $SL_{2}(A)$, we have that $w$ acts on $T$ by conjugation.
My doubt is to know how the induced morphism in the first homology by the conjugation $w_{1}:H_{1}(T,\mathbb{Z})\rightarrow H_{1}(T,\mathbb{Z})$ is the map $u\mapsto u^{-1}$.
I was trying to do it by using the isomorphism $T=T^{ab}\cong H_{1}(T,\mathbb{Z})\rightarrow H_{1}(T,\mathbb{Z})\cong T^{ab}=T$ which I can get the result by using the bar resolution and the isomorphism $A^{\ast}\cong T(A)$, however I was told that by using the functoriality of the first homology is enough without using direct calculations.
The only thing I think is that the isomorphism $H_{1}(G,\mathbb{Z})\cong G^{ab}$ is natural.
if the latter is true, it is enough by using the conjugation map $t_{w}:T\rightarrow T$ which is only inversion in this particular case. I would like to know if that is correct.
Thanks!