In Spivak’s Calculus on Manifolds, pg. 69, he claims that, if we define $T = Dg(a)$, then $(T^{-1}\circ g)’ (a) = I$, where $I$ is the identity. Using the inverse function theorem, I am getting
$(T^{-1}\circ g)’ (a) = (T^{-1}\circ T’)^{-1})\circ g(a) \cdot g’(a)$,
after which I get
$(T’)^{-1}\circ T \circ g(a) \cdot g’(a)$.
I believe $(T’)^{-1}$ is the identity, considering this $T$ is a constant matrix, but I can’t make anymore progress.
$T$ is a fixed linear map. The derivative of any linear map is itself (at any point of the vector space). Go back to the definition if you're not sure about this. You should have $$(T^{-1})'(g(a))\cdot g'(a) = T^{-1}\cdot g'(a) = (g'(a))^{-1}\cdot g'(a) = I.$$