Lemma 2.0.12 (Gluing for Linear Functions). Let $E, F$ be vector spaces over a field $K$, $\fF$ be a family of subspaces of $E$, and $\bracs{T_V}_{V \in \fF}$ with $T_{V} \in \hom(V; F)$ for all $V \in \fF$. If:

  1. $\bigcup_{V \in \fF}V = E$.

  2. For each $V, W \in \fF$, $T_{V}|_{V \cap W}= T_{W}|_{V \cap W}$.

  3. $\fF$ is upward-directed with respect to includion.

then there exists a unique $T \in \hom(E; F)$ such that $T|_{V}= T_{V}$ for all $V \in \fF$.

Proof. By (a), (b), and Lemma 2.0.11, there exists a unique $T: E \to F$ such that $T|_{V} = T_{V}$ for all $V \in \fF$.

Let $x, y \in E$ and $\lambda \in \fF$. By assumption (a), there exists $V_{x}, V_{y} \in \fF$ with $x \in V_{x}$ and $y \in V_{y}$. By assumption (3), there exists $V \in \fF$ with $V \supset V_{x} \cup V_{y}$. Hence

\[T(\lambda x + y) = T_{V}(\lambda x + y) = \lambda T_{V}x + T_{V}y = \lambda Tx + Ty\]

and $T \in \hom(E; F)$.$\square$