\chapter{Gluing Lemmas}
\label{chap:gluing}


\begin{lemma}[Gluing for Functions]
\label{lemma:glue-function}
    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:
    \begin{enumerate}
        \item[(a)] $\bigcup_{i \in I}U_i = X$.
        \item[(b)] For each $i, j \in I$, either $U_i \cap U_j = \emptyset$, or $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$.
    \end{enumerate}
    then there exists a unique $f: X \to Y$ such that $f|_{U_i} = f_i$ for all $i \in I$.
\end{lemma}
\begin{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:
    \begin{enumerate}
        \item By assumption (a), $\bracs{x|(x, y) \in \Gamma} = \bigcup_{i \in I}U_i = X$.
        \item 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), $y = y'$.
    \end{enumerate}
    Thus $\Gamma$ is the graph of a function $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
\end{proof}

\begin{lemma}[Gluing for Linear Functions]
\label{lemma:glue-linear}
    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:
    \begin{enumerate}
        \item[(a)] $\bigcup_{V \in \fF}V = E$.
        \item[(b)] For each $V, W \in \fF$, $T_V|_{V \cap W} = T_W|_{V \cap W}$.
        \item[(c)] $\fF$ is upward-directed with respect to includion.
    \end{enumerate}
    then there exists a unique $T \in \hom(E; F)$ such that $T|_{V} = T_V$ for all $V \in \fF$.
\end{lemma}
\begin{proof}
    By (a), (b), and \autoref{lemma:glue-function}, 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
    \[
    T(\lambda x + y) = T_V(\lambda x + y) = \lambda T_Vx + T_Vy = \lambda Tx + Ty
    \]

    and $T \in \hom(E; F)$.
\end{proof}
