= Solution
\b[True.] Choose a <basis> $v_1,\ldots,v_r$ of $V$. By <surjective function>, choose $u_i\in U$ with $\alpha u_i=v_i$. Define the <linear map> $\beta$ by $\beta v_i=u_i$ and extend by <linearity>. Then $\alpha\beta v_i=v_i$ on every <basis> vector, and therefore
$$
\boxed{\alpha\beta=\operatorname{id}_V}.
$$
This constructs a <right inverse> and does not require $\alpha$ to be <injective>. If $V$ is the zero <vector space>, the unique zero map is the required <right inverse>.
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