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I am stuck halfway through this problem, and I feel like it should be easier than I am making it. :(

I need to prove that $A^{B×C}$ and $(A^B)^C$ are a bijection where A, B and C are all sets.

Anyway, this is what I have so far.

$g \in A^{B×C}$ means that $g(b,c)\in A$ when $b \in B$ and $c\in C$.

$f\in (A^B)^C$ means that $f(c)\in A$ when $c\in C$, which in turn means that $f(c)(b)\in A$ when $b \in B$.

I need to find some sort of function $G:A^{B×C} \to (A^B)^C$.

babayaga
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    I've edited your post to make the mathjax look the way I think you wanted - the command \in is used for $\in$ and curly braces let you put a group in the exponent, I used A^{B×C} – B. Mehta May 18 '17 at 18:25
  • Thank you! I'm still new here. – babayaga May 18 '17 at 18:25
  • That's alright! In the future, this reference guide may help: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – B. Mehta May 18 '17 at 18:27
  • This is currying: https://en.wikipedia.org/wiki/Currying – Angina Seng May 18 '17 at 18:50

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Let $G(g)(c)(b)=g(b,c)$. That is: Given $g\colon B\times C\to A$ we need that $G(g)$ is a function from $C$ to $A^B$, i.e., we specify $G(g)$ by specifying for each $c\in C$ a map $B\to A$. That map $B\to A$ is simply $g(\cdot,c)$.

Hagen von Eitzen
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  • I'm a first-year student and easily confused. Why would a map like that help me?

    $A^{B×C}$ is the set of all functions from an ordered pair $B×C$ to a number $A$. But $(A^B)^C$ is the set of all functions from C to the set of all functions from B to A, so it's the set of all functions from a number (in C) to a function.

    One is functions from pair to number. The other is functions from number to function. Wouldn't I want compose them to get something from pair to number to function? If not, why not?

     – babayaga May 18 '17 at 19:34