Let \(a,b,c,d\in\CC\) with \(ad-bc\neq0\). The linear fractional transformation with coefficients \(a,b,c,d\) is \[ T(z)\coloneqq\frac{az+b}{cz+d}, \] defined for \(z\in\CC\) with \(cz+d\neq0\).
The condition \(ad-bc\neq0\) says that the vectors \((a,b)\) and \((c,d)\) are linearly independent. In particular \((c,d)\neq(0,0)\), so the domain \(\{z\in\CC:cz+d\neq0\}\) is \(\CC\) minus at most one point, and \(T\) is not constant there.
Let \(T\) be the linear fractional transformation with coefficients \(a,b,c,d\), where \(ad-bc\neq0\). Then \(T\) is a bijection from \(\{z\in\CC:cz+d\neq0\}\) onto \(\{w\in\CC:-cw+a\neq0\}\), with inverse the linear fractional transformation \[ T^{-1}(w)=\frac{dw-b}{-cw+a}. \]
Let \(S(w)\coloneqq(dw-b)/(-cw+a)\); its coefficients satisfy \(da-(-b)(-c)=ad-bc\neq0\). For \(cz+d\neq0\), \[ -c\,T(z)+a=\frac{-c(az+b)+a(cz+d)}{cz+d}=\frac{ad-bc}{cz+d}\neq0, \] so \(T(z)\) lies in the domain of \(S\), and \[ S(T(z))=\frac{d(az+b)-b(cz+d)}{-c(az+b)+a(cz+d)}=\frac{(ad-bc)z}{ad-bc}=z. \] The same computation with the roles of \(T\) and \(S\) exchanged gives \(c\,S(w)+d\neq0\) and \(T(S(w))=w\) for \(-cw+a\neq0\).