= Solution
{c}
Let $F$ be the common <field>. The <dual map> is $S'(f)=f\circ S$ for $f\in W'$, where $W'$ and $V'$ are the <dual spaces>.
First assume $S'$ is <injective>. If $S$ were not <surjective>, its <image of a linear map> $\operatorname{im}S$ would be a <proper vector subspace> of $W$. Choose a <basis> $w_1,\ldots,w_r$ of $\operatorname{im}S$ and use <basis extension> to obtain a <basis> $w_1,\ldots,w_m$ of $W$, with $r<m$. There is a <linear functional> $f\in W'$ with
$$
f(w_{r+1})=1,\qquad f(w_j)=0\quad(j\ne r+1).
$$
It is nonzero and vanishes on $\operatorname{im}S$, so $S'(f)=f\circ S=0$, contradicting <injectivity>. Equivalently, the <annihilator of a vector subspace> $\operatorname{im}S$ would give a nonzero <kernel> for $S'$. Therefore
$$
\boxed{S'\text{ injective}\ \Longrightarrow\ S\text{ surjective}.}
$$
Now assume $S'$ is <surjective>. If $0\ne v\in\ker S$, <basis extension> gives a <linear functional> $h\in V'$ with $h(v)=1$: include $v$ in a <basis> and take its <coordinate functional>. By <surjectivity>, $h=S'(f)$ for some $f\in W'$. But
$$
1=h(v)=S'(f)(v)=f(Sv)=f(0)=0,
$$
a contradiction. Hence $\ker S=0$, which proves
$$
\boxed{S'\text{ surjective}\ \Longrightarrow\ S\text{ injective}.}
$$
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