= Solution
The dual space is
$$
V^*=\operatorname{Hom}_F(V,F),
$$
the <vector space> of linear functionals on $V$. If $e_1,\ldots,e_n$ is a <basis> of a finite-dimensional $V$, define $e_i^*(e_j)=\delta_{ij}$. Every $f\in V^*$ has the unique expansion
$$
f=\sum_{i=1}^nf(e_i)e_i^*,
$$
so the <dual basis> $e_1^*,\ldots,e_n^*$ proves
$$
\boxed{\dim V^*=\dim V=n}.
$$
For $U\leq V$, its <annihilator of a vector subspace> is
$$
U^\circ=\{f\in V^*:f(u)=0\text{ for every }u\in U\}.
$$
If $u_1,\ldots,u_k$ is a <basis> of $U$ and is extended to a <basis> $u_1,\ldots,u_n$ of $V$, then
$$
U^\circ=\operatorname{span}\{u_{k+1}^*,\ldots,u_n^*\}.
$$
Consequently
$$
\boxed{\dim U^\circ=\dim V-\dim U}.
$$
If $U\ne V$, this dimension is positive, giving a nonzero functional that vanishes on $U$.
For a <linear map> $\alpha:V\to W$, the <dual map> is
$$
\alpha^*:W^*\to V^*,
\qquad
\alpha^*(g)=g\circ\alpha.
$$
Now
$$
g\in\ker\alpha^*
\Longleftrightarrow g(\alpha v)=0\text{ for every }v
\Longleftrightarrow g\in(\operatorname{im}\alpha)^\circ,
$$
so
$$
\boxed{\ker\alpha^*=(\operatorname{im}\alpha)^\circ}.
$$
Every $g\circ\alpha$ vanishes on $\ker\alpha$, hence
$$
\operatorname{im}\alpha^*\subseteq(\ker\alpha)^\circ.
$$
The two spaces have the same dimension, since
$$
\begin{aligned}
\dim\operatorname{im}\alpha^*
&=\dim W-\dim(\operatorname{im}\alpha)^\circ
=\dim\operatorname{im}\alpha,\\
\dim(\ker\alpha)^\circ
&=\dim V-\dim\ker\alpha
=\dim\operatorname{im}\alpha.
\end{aligned}
$$
Thus
$$
\boxed{\operatorname{im}\alpha^*=(\ker\alpha)^\circ}.
$$
Let $Q:V\to V/U$ be the quotient map. Since $Q$ is onto, $Q^*$ is injective, and the preceding identity gives
$$
\operatorname{im}Q^*=(\ker Q)^\circ=U^\circ.
$$
Therefore
$$
\boxed{(V/U)^*\cong U^\circ}.
$$
For the inclusion $J:U\hookrightarrow V$, the map $J^*:V^*\to U^*$ is restriction to $U$. It is onto and has kernel
$$
\ker J^*=(\operatorname{im}J)^\circ=U^\circ.
$$
The first isomorphism theorem now gives the other <duals of a subspace and its quotient>:
$$
\boxed{U^*\cong V^*/U^\circ}.
$$
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