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:
<br /> f_1:\B \rightarrow A<br />

<br /> f_2:\C \rightarrow A<br />

<br /> f=f_1\cup f_2<br />

(2) Consider the application:

<br /> \Phi:A^{B+C}\rightarrow A^{B} A^{C}<br />

Defined by:

<br /> \Phi\left(f\right)\left(a\right) = \left(f_1\left(a\right),f_2\left(a\right)\right)<br />

For a \in A. Is it a bijection?