Let $A$ be an $n\times m$ matrix with rank $n$. Then I would like to conclude that there is a matrix $B$ such that $AB=I_n$.
This is easy, on the intuitive level. The image of $A$ in $\mathbb{R}^m$ is an $n$-dimensional subspace, spanned by the pivotal columns of $A$. Taking $B$ to be the $m\times n$ matrix that sends each pivotal column to the standard basis in $\mathbb{R}^n$ should yield the answer.
I feel like my solution lacks rigor. Any tips for how to make this more precise?