= Solution
{c}
The <quotient vector space> $V/U$ identifies two <vectors> $v,w$ precisely when $v-w\in U$. Since $U$ is an <invariant subspace>, $T(U)\subseteq U$. Thus
$$
v+U=w+U\ \Longrightarrow\ v-w\in U\ \Longrightarrow\ T(v)-T(w)=T(v-w)\in U.
$$
Consequently $T(v)+U=T(w)+U$: the output does not depend on the chosen representative. This proves that the <induced linear map on a quotient vector space>
$$
\boxed{\overline T(v+U)=T(v)+U}
$$
is well-defined. It is a <linear map> because, for <scalars> $a,b$,
$$
\overline T\bigl(a(v+U)+b(w+U)\bigr)
=T(av+bw)+U=a\overline T(v+U)+b\overline T(w+U).
$$
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