Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 1 / c / iii / Solution

The necessary and sufficient condition isFor necessity, if $ST={\mathrm{id}}_V$ and $v\in \ker T$, then $v=S(Tv)=S(0)=0$.

For sufficiency, let $e_1,e_2$ be a basis of $V$ and suppose the kernel of a linear map is trivial. The vectors $T{e}_1,T{e}_2$ are linearly independent, since $aT{e}_1+bT{e}_2=0$ implies $a{e}_1+b{e}_2\in \ker T$. Extend them to a basis $T{e}_1,T{e}_2,w_3$ of $W$. Define a linear map $S$ byThen $ST$ agrees with the identity operator on a basis, hence on all of $V$. Thus a left inverse exists precisely when $T$ is injective; this is the linear left inverse criterion.