Prove A^B A^C = A^{B+C}: Steps & Tips

[Deadstar]

Let A,B,C be three cardinals. Show that...

$$A^{B+C} = A^B A^C$$

I thought about using some some of distinguishing scheme where we denoted $$B \cup C$$ as $$B \times \{0\} \cup C \times \{1\}$$ so we could map thing easily but apparently that's not right and you can assume they are disjoint anyway...

EDIT: I just noticed the sticky at the top, this was just something in the notes we had that wasn't proven, not an assignment question or anything and I wanted to know how to do it. Should I repost this in another section?

 

[JSuarez]

Two hints:

(1) Assuming that B and C are disjoint (which we may always assume by that construction, because it doesn't change the cardinality), any function from B + C (the disjoint union of B and C) may always be thought as the union of two functions:
$$f_1:\B \rightarrow A$$

$$f_2:\C \rightarrow A$$

$$f=f_1\cup f_2$$

(2) Consider the application:

$$\Phi:A^{B+C}\rightarrow A^{B} A^{C}$$

Defined by:

$$\Phi\left(f\right)\left(a\right) = \left(f_1\left(a\right),f_2\left(a\right)\right)$$

For $a \in A$. Is it a bijection?