If ${I,T \in B(V,V)}$ such that ${(I-T)}$ is invertible - is ${(I-T)^{-1}-I^{-1} = (I-T)^{-1}\circ T}$?

Question

I just wanted to make sure I haven't done anything stupid here. ${I^{-1}=I}$ since $I$ is the identity map, and because of linearity $$ (I-T)^{-1} - I^{-1} = (I-T)^{-1}\circ (I - (I-T)) = (I-T)^{-1}\circ T $$ Does this make sense? Thanks!