When a union of a family $\{f_i \}_{i \in I}$ of functions $f_i: X_i \to Y_i$ is itself a function $\bigcup_{i \in I} X_i \to \bigcup_{i \in I} Y_i$?

Question

I wish to know when a union of a family $\{f_i \}_{i \in I}$ of functions $f_i: X_i \to Y_i$ is itself a function $\bigcup_{i \in I} X_i \to \bigcup_{i \in I} Y_i$.

I take the definition of a function $X \to Y$ as of a relation $f \subseteq X \times Y$ so that $\mathrm{dom \ f} = \{ x \in X \ | \ \exists y \in Y: \ (x,y) \in f \} = X$ and $(x,y_1), \ (x,y_2) \in f \Rightarrow y_1 = y_2$.

Now, let $\{ f_i \}_{i \in I}$ be a family of functions, $f_i \in {Y_i}^{X_i}$.

I know that $\bigcup_{i \in I} f_i \subseteq \bigcup_{i \in I} X_i \times Y_i \subseteq \bigcup_{i \in I} X_i \times \bigcup_{i \in I} Y_i$.

What is more, $x \in \bigcup_{i \in I} X_i \Leftrightarrow \exists i \in I: \ x \in X_i \Rightarrow \exists y \in Y_i: \ (x,y) \in f_i \Rightarrow \exists y \in \bigcup_{i \in I} Y_i: (x,y) \in \bigcup_{i \in I} f_i \Leftrightarrow x \in \mathrm{dom \ \bigcup_{i \in I} f_i}$.

What is left to check is when $(x,y_1), (x,y_2) \in \bigcup_{i \in I} f_i \Rightarrow y_1 = y_2$.

But $(x,y_1), (x,y_2) \in \bigcup_{i \in I} f_i$ means that there exists $i, j \in I$ so that $f_i(x) = y_1$ and $f_j(x) = y_2$. Is it right?

Answer 1

You are right. For the union to be a function you therefore need that $f_i$ and $f_j$ agree on $X_i \cap X_j$ for all $i,j$. Hence then you can continue, with $y_1 = f_i(x) = f_j(x) = y_2$ are conclude that $\bigcup_i f_i$ is a function.

So the searched condition is: $$ f_i \cap ((X_i \cap X_j) \times Y_i) = f_j \cap ((X_i \cap X_j)\times Y_j) $$