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Let us consider $ f | _B $ the function restricted to $ B $ and analogously $ f | _C $. The formula for both is $ f | _B (b) = f (b) $ and $ f | _C (c) = f (c) $, for all $ b, c \ in B, C $ respectively.

Let's define $ F: A ^ {B \cup C} \to A ^ B \times A ^ C $, given by $$ F (f) = (f | _B, f | _C) $$.

To demonstrate injectivity, let $ f, f '\in A ^ {B \cup C} $, such that $ F (f) = F (f') $, that is $ (f | _B, f | _C) = (f '| _B, f' | _C) $. Hence, it follows that $ f (b) = f | _B = f '| _B = f' (b) $ and $ f (c) = f | _C = f '| _C = f' (c) $. This is that $ f = f '$. That way $ F $ is injective.

Some help demonstrating surjectivity.

James A.
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  • If $B=C$ then the left hand is $A^B$ but the right hand is $A^B\times A^B$. – lulu Feb 15 '21 at 21:26
  • I forgot to put that $ B \cap C = \varnothing $. – James A. Feb 15 '21 at 21:28
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    But if $B\cap C=\emptyset$ then, given any $g:C\to A$ and any $h:B\to A$ we can define $f:B\cup C\to A$ by $f(x)=g(x)$ if $x\in C$ and $f(x)=h(x)$ if $x\in B$. That's your inverse map. Of course, without the assumption on the intersection, that map $f$ would not be well defined. – lulu Feb 15 '21 at 21:31
  • I know, you are right. My mistake was not putting this missing hypothesis. Thank you very much for that help. – James A. Feb 15 '21 at 21:34

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In general, given sets $B$ and $C$, we can form the disjoint union $B \sqcup C$ by choosing disjoint copies $B'$ of $B$ and $C'$ of $C$, and then $B \sqcup C = B' \cup C'$; for example $$B \sqcup C := (B \times \{0\}) \cup (C \times \{1\}).$$ Now, we claim that there is a bijection between $A^{B \sqcup C}$ and $A^B \times A^C$ (this is the reason of why the disjoint union $B \sqcup C$ is also denoted by $B + C$, or $B \oplus C$). Indeed, given functions $g: B \to A$ and $h : C \to A$, define $g \sqcup h : B \sqcup C \to A$ by $$(g \sqcup h) \circ j_B = g \textrm{ and } (g \sqcup h) \circ j_C = h$$ where $j_Z : Z \to B \sqcup C$ (for $Z \in \{B,C\}$) sends each element in $Z$ to its corresponding copy in $Z' \subseteq B \sqcup C$. Thus, the function $$A^B \times A^C \to A^{B \sqcup C}; \quad (g,h) \mapsto g \sqcup h$$ is a bijection, with inverse $$A^{B \sqcup C} \to A^B \times A^C; \quad f \mapsto (f \circ j_B,f \circ j_C).$$

azif00
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