Find a bijection from $(A^B)^C$ into $A^{B \times C}$

Question

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How to show $(a^b)^c=a^{bc}$ for arbitrary cardinal numbers?

Notation: Let A and B be sets. The set of all functions $f:A \rightarrow B$ is denoted by $B^A$.

Problem: Let A, B, and C be sets. Show that there exists a bijection from $(A^B)^C$ into $A^{B \times C} $. You should first construct a function and then prove that it is a bijection.

What I have so far:

$f:C \rightarrow A^B$

$ f(c) : B \rightarrow A $

and

$ g : B \times C \rightarrow A $

$ g(b, c) \in A$

How do I find a bijection between these functions?

score 3 · Accepted Answer · answered Aug 03 '12 at 05:15

3

Let $f \in (A^B)^C, g \in A^{B \times C}$. Define $\Phi: (A^B)^C \to A^{B \times C}$ by setting

$$\Phi(f)(b,c) = f(b)(c)$$

This is a bijection because it has an inverse $\Psi: A^{B \times C} \to (A^B)^C$

$$\Psi(g)(b)(c) = g(b,c)$$

answered Aug 03 '12 at 05:15

Elchanan Solomon

30,807

I'm unfamiliar with the notation $ f(b)(c) $. Does this imply a composition? – BrandonK. Aug 03 '12 at 05:16
3

@BrandonK. It denotes two evaluations: $f$ is a function-valued function, so $f(b)$ is itself a function, so it can be evaluated at $c$ to get $f(b)(c)$. – Zhen Lin Aug 03 '12 at 05:24
Thanks, it makes a lot more sense now. – BrandonK. Aug 03 '12 at 05:30

Find a bijection from $(A^B)^C$ into $A^{B \times C}$

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