= Solution
{c}
The necessary and sufficient condition is
$$
\boxed{\ker T=\{0\}.}
$$
For necessity, if $ST=\operatorname{id}_V$ 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> $Te_1,Te_2$ are <linearly independent>, since $aTe_1+bTe_2=0$ implies $ae_1+be_2\in\ker T$. Extend them to a <basis> $Te_1,Te_2,w_3$ of $W$. Define a <linear map> $S$ by
$$
S(Te_1)=e_1,\qquad S(Te_2)=e_2,\qquad S(w_3)=0.
$$
Then $ST$ 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>.
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