Answer
$T^{-1}$ maps $\mathrm{R}^{n}$ onto $\mathrm{R}^{n}$ and
$T^{-1}$ is a one-to-one mapping of $\mathrm{R}^{n}$ onto $\mathrm{R}^{n}$
Work Step by Step
Let $T$ be such that $T$ maps $\mathrm{R}^{n}$ onto $\mathrm{R}^{n}$. Let the standard matrix of $\mathrm{T}$ be $A$.
By Th.12.a of Chapter 1, the columns of its standard matrix $A$ span $\mathrm{R}^{n}$.
By the IMT,
$\mathrm{a}. \quad A$ is an invertible matrix.
and
$\mathrm{h}.\ \quad$The columns of $A$ span $\mathbb{R}^{n}$.
are both true, so $A$ is invertible.
By Th.9, $T$ is invertible, and the standard matrix of $\mathrm{T}^{-1}$ is $\mathrm{A}^{-1}.$
Again, by the IMT, this time since $\mathrm{A}^{-1}$ is invertible (a),
the columns of $\mathrm{A}^{-1}$ are linearly independent (e) and span $\mathrm{R}^{n}$ (h.)
By Th.12 of Chapter 1, applied on $T^{-1}$:
$T^{-1}$ maps $\mathrm{R}^{n}$ onto $\mathrm{R}^{n}$ and,
since the columns of $A^{-1}$ are linearly independent,
$T^{-1}$ is a one-to-one mapping of $\mathrm{R}^{n}$ onto $\mathrm{R}^{n}$.
