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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?

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marked as duplicate by Zhen Lin, yunone, Asaf Karagila, Brandon Carter, J. M. Aug 3 '12 at 8:09

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1 Answer 1

up vote 3 down vote accepted

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 $\Psi: A^{B \times C} \to (A^B)^C$

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

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I'm unfamiliar with the notation $ f(b)(c) $. Does this imply a composition? –  BrandonK. Aug 3 '12 at 5:16
2  
@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 3 '12 at 5:24
    
Thanks, it makes a lot more sense now. –  BrandonK. Aug 3 '12 at 5:30

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