= Solution
{c}
Write $K_j=\ker T^j$, where $T^0=I$ and $K_0=\{0\}$. These <kernels> satisfy $K_j\subseteq K_{j+1}$. Also $T(K_{i+1})\subseteq K_i$ and $T(K_i)\subseteq K_{i-1}$: for example, $T^i(Tv)=T^{i+1}v=0$ when $v\in K_{i+1}$. Thus the proposed <linear map> between <quotient vector spaces> is well-defined and has the formula
$$
\overline T:K_{i+1}/K_i\longrightarrow K_i/K_{i-1},\qquad
v+K_i\longmapsto Tv+K_{i-1}.
$$
If $v+K_i$ belongs to its <kernel>, then $Tv\in K_{i-1}$. Therefore $T^iv=T^{i-1}(Tv)=0$, so $v\in K_i$ and $v+K_i$ is zero. We have proved
$$
\boxed{\overline T\text{ is injective for every }i\geq1.}
$$
These are the <injections between successive kernel quotients>. In finite <dimension>, they imply $\dim K_{i+1}-\dim K_i\leq\dim K_i-\dim K_{i-1}$.
\b[Surjectivity need not hold.] Take a nonzero <vector space> $V$ and the zero <linear map> $T=0$. At $i=1$, we have $K_2=K_1=V$ and $K_0=0$, so
$$
\boxed{\overline T:V/V\longrightarrow V/0\text{ is the map }0\longrightarrow V,\text{ which is not surjective}.}
$$
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