2 Gluing Lemmas
Lemma 2.0.11 (Gluing for Functions). Let $X, Y$ be sets, $\seq{U_i}\subset 2^{X}$, and $\seqi{f}$ with $f_{i}: U_{i} \to Y$ for all $i \in I$. If:
$\bigcup_{i \in I}U_{i} = X$.
For each $i, j \in I$, either $U_{i} \cap U_{j} = \emptyset$, or $f_{i}|_{U_i \cap U_j}= f_{j}|_{U_i \cap U_j}$.
then there exists a unique $f: X \to Y$ such that $f|_{U_i}= f_{i}$ for all $i \in I$.
Proof. For each $i \in I$, let $\Gamma_{i} \subset U_{i} \times Y$ be the graph of $f_{i}$. Let $\Gamma = \bigcup_{i \in I}\Gamma_{i}$, then:
By assumption (a), $\bracs{x|(x, y) \in \Gamma}= \bigcup_{i \in I}U_{i} = X$.
For any $x \in X$, there exists $y \in Y$ with $(x, y) \in \Gamma$, and $i \in I$ such that $(x, y) \in \Gamma_{i}$. If $(x, y') \in \Gamma_{j} \subset \Gamma$, then $x \in U_{i} \cap U_{j} \ne \emptyset$. By assumption (b), $y = y'$.
Thus $\Gamma$ is the graph of a function $f: X \to Y$ with $f|_{U_i}= f_{i}$ for all $i \in I$.$\square$
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:
$\bigcup_{V \in \fF}V = E$.
For each $V, W \in \fF$, $T_{V}|_{V \cap W}= T_{W}|_{V \cap W}$.
$\fF$ 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
and $T \in \hom(E; F)$.$\square$