Proof of Theorem KPI

Let $M=\setparts{\vect{u}+\vect{z}}{\vect{z}\in\krn{T}}$. First, we show that $M\subseteq\preimage{T}{\vect{v}}$. Suppose that $\vect{w}\in M$, so $\vect{w}$ has the form $\vect{w}=\vect{u}+\vect{z}$, where $\vect{z}\in\krn{T}$. Then \begin{align*} \lt{T}{\vect{w}}&=\lt{T}{\vect{u}+\vect{z}}\\ &=\lt{T}{\vect{u}}+\lt{T}{\vect{z}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\vect{v}+\zerovector&&\vect{u}\in\preimage{T}{\vect{v}},\ \vect{z}\in\krn{T}\\ &=\vect{v}&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}} \end{align*} which qualifies $\vect{w}$ for membership in the preimage of $\vect{v}$, $\vect{w}\in\preimage{T}{\vect{v}}$.

For the opposite inclusion, suppose $\vect{x}\in\preimage{T}{\vect{v}}$. Then, \begin{align*} \lt{T}{\vect{x}-\vect{u}}&=\lt{T}{\vect{x}}-\lt{T}{\vect{u}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\vect{v}-\vect{v}&&\vect{x},\,\vect{u}\in\preimage{T}{\vect{v}}\\ &=\zerovector \end{align*}

This qualifies $\vect{x}-\vect{u}$ for membership in the kernel of $T$, $\krn{T}$. So there is a vector $\vect{z}\in\krn{T}$ such that $\vect{x}-\vect{u}=\vect{z}$. Rearranging this equation gives $\vect{x}=\vect{u}+\vect{z}$ and so $\vect{x}\in M$. So $\preimage{T}{\vect{v}}\subseteq M$ and we see that $M=\preimage{T}{\vect{v}}$, as desired.