= Solution
A <linear map> preserves <linear combinations>: $\psi(av+bw)=a\psi(v)+b\psi(w)$ for all vectors and real scalars. The <rank-nullity theorem> gives $\dim V=\dim\ker\psi+\dim\operatorname{im}\psi$. For an endomorphism of the same finite-dimensional <vector space>, surjectivity means rank $\dim V$, equivalent to zero kernel and hence injectivity.
If $\psi\phi=I$, then every $v=\psi(\phi v)$ lies in the image, so $\psi$ is surjective and consequently injective. Now $\psi(\phi\psi v)=\psi v$; injectivity implies $\phi\psi v=v$ for every $v$. Thus \b[the right inverse is also a left inverse:] $\boxed{\phi\psi=I}$. This is the <one-sided inverses of finite-dimensional endomorphisms> property.
Apply it to the <linear maps> on $\mathbb R^n$ represented by $A$ and $B$: $AB=I$ implies $\boxed{BA=I}$. Equal finite dimensions are essential to the argument.
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