Proposition 7

(1) Let E be a set and $(X_\iota)_{\iota \in I}$ a covering of E. If $f$ and $g$ are two functions with domain E such that $f$ and $g$ agree on $E \cap X_\iota$ for each $\iota \in I$, then $f$ and $g$ agree on E.

(2) Let $(X_\iota)_{\iota \in I}$ be a family of sets and let $(f_\iota)_{\iota \in I}$ be a family of mappings with the same target F such that for each $\iota \in I$ the domain of $f_\iota$ is $X_\iota$, and for each pair $(\iota, \varkappa) \in I \times I$, $f_\iota$ and $f_\varkappa$ agree on $X_\iota \cap X_\varkappa$. Then there is exactly one function $f$ with domain $A = \bigcup_{\iota \in I} X_\iota$ and target F which extends each of the functions $f_\iota$ $(i \in I)$.

(1) Let $x$ be any element of E. Then there exists $\iota \in I$ such that $x \in X_\iota$, whence $f(x) = g(x)$ by hypothesis. (2) Let $G_\iota$ be the graph of $f_\iota$ and let $G = \bigcup_{\iota \in I} G_\iota$; let us show that G is a functional graph. If $(x, y) \in G$ and $(x, y') \in G$, there exist two indices $\iota$, $\varkappa$ in I such that $(x, y) \in G_\iota$ and $(x, y') \in G_\varkappa$. This implies $x \in X_\iota$, $x \in X_\varkappa$, $y = f_\iota(x)$, $y' = f_\varkappa(x)$; but since $x \in X_\iota \cap X_\varkappa$, we have

that is to say, $y = y'$. The graph G has domain $\mathrm{pr}_1 G = \bigcup_{\iota \in I} \mathrm{pr}_1 G_\iota = A$; the function $f = (G, A, F)$ therefore satisfies the required conditions. Its uniqueness follows from the first part of the Proposition.