= Solution
\b[True.] Use the lifted <basis> $u_1,\ldots,u_r$ above and put $W=\operatorname{span}\{u_1,\ldots,u_r\}$. If $\sum c_i u_i=0$, applying $\alpha$ gives $\sum c_i v_i=0$, so all $c_i=0$. Thus the $u_i$ form a <basis> of $W$, and $\alpha|_W$ takes this <basis> to the <basis> of $V$. It is a <linear isomorphism>.
In fact the <right inverse> gives the stronger splitting
$$
\boxed{U=\ker\alpha\oplus W}.
$$
For every $u\in U$, write $u=(u-\beta\alpha u)+\beta\alpha u$; the first term is in the <kernel> and the second in $W$. Their intersection is zero because $\alpha|_W$ is <injective>. This is a <direct sum> decomposition.
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