= Solution
The <annihilator of a vector subspace> is
$$
U^\circ=\{\phi\in V':\phi(u)=0\text{ for every }u\in U\}.
$$
Choose a <basis> $e_1,\ldots,e_r$ of $U$ and extend it to a <basis> $e_1,\ldots,e_d$ of $V$. In the <dual basis>, precisely $f_{r+1},\ldots,f_d$ span $U^\circ$. Therefore
$$
\boxed{\dim U^\circ=\dim V-\dim U.}
$$
The canonical restriction <linear transformation>
$$
\rho:V'\longrightarrow U',\qquad \rho(\phi)=\phi|_U
$$
has <kernel of a linear map> $U^\circ$. It is surjective: extend any <linear functional> on $U$ by assigning arbitrary values, for instance zero, to the additional <basis> vectors. The <First isomorphism theorem for vector spaces> gives the explicit <linear isomorphism>
$$
\boxed{V'/U^\circ\ \longrightarrow U',\qquad
\phi+U^\circ\longmapsto\phi|_U.}
$$
It is natural because the displayed map itself makes no <basis> choice; a <basis> was used only to prove surjectivity. In particular, $U'$ is canonically a <quotient vector space> of $V'$, rather than canonically a subspace of $V'$.
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