= Solution
{c}
The <preimage> is
$$
T^{-1}(X)=\{u\in U:T(u)\in X\}.
$$
It contains $0$, and if $u,v$ belong to it and $a,b\in\mathbb R$, then $T(au+bv)=aT(u)+bT(v)\in X$, since $X$ is a <vector subspace>. Thus $T^{-1}(X)$ is a <vector subspace>. It contains the <kernel> of $T$ because $0\in X$.
To establish the <dimension of the preimage of a vector subspace> formula without assuming the theorem to be deduced, choose a <basis> $k_1,\ldots,k_{n_T}$ of $\ker T$ and a <basis> $y_1,\ldots,y_d$ of $X\cap\operatorname{im}T$, where $d=\dim(X\cap\operatorname{im}T)$. For each $j$, choose $u_j\in U$ with $T(u_j)=y_j$. These $u_j$ lie in $T^{-1}(X)$.
The family $k_1,\ldots,k_{n_T},u_1,\ldots,u_d$ is <linearly independent>: applying $T$ to a vanishing <linear combination> forces all $u_j$ <coefficients> to vanish by <linear independence> of the $y_j$, and then all remaining <coefficients> vanish by <linear independence> of the $k_i$. It also <spans> $T^{-1}(X)$: for $u$ in that <preimage>, write $T(u)=\sum_j b_jy_j$. Then $u-\sum_j b_ju_j\in\ker T$, so it is a <linear combination> of the $k_i$. Consequently
$$
\boxed{\dim T^{-1}(X)=\dim(X\cap\operatorname{im}T)+n_T.}
$$
Set $X=V$. Then $T^{-1}(V)=U$ and $V\cap\operatorname{im}T=\operatorname{im}T$, giving the <rank-nullity theorem>:
$$
\boxed{\dim U=r_T+n_T.}
$$
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