2

I'm having trouble proving that $|A^{B×C}| = |(A^B)^C|$ , where $M^N$ is the set of all the functions $f:N \to M$.

My thoughts: to prove this, I need to find a bijection between $|A^{B×C}|$ and $|(A^B)^C|$, so I need a bijection between the set of all functions $g:B×C \to A$ and the set of all functions $h: C \to A^B$.

Lstoi
  • 199

2 Answers2

2

The bijection actually should be between the set of all functions $g:B\times C \rightarrow A$ and the set of all functions $h:C\rightarrow A^B$ (otherwise you'd be proving that $|A^{B\times C}|=|A^C|$, which is false in general).

Added after the asker's edit:

In the wikipedia article of currying linked by Clive Newstead, there is a mention of the important and quite general phenomenon of "adjointness". Let's change the notation a little bit: Put $\hom(B,A)=A^B$. Then we are looking for bijections

$\hom(B\times C,A)\cong\hom(C\times B,A)\cong \hom(C,\hom(B,A))$.

This is quite helpful when trying to remember how does the explicit bijection comes around: for a function $f:C\times B\rightarrow A$, fixing a coordinate $c_0\in C$ will yield a function $f(c_0,-):B\rightarrow A$, so $f$ can be seen as a function such that, when evaluated in a point $c_0$, yields a function from $B$ to $A$ (this of course is not a proof, rather a hint).

Aldo Guzmán Sáenz
  • 561
1

First, an element of $A^{B\times C}$ is a function $f:B\times C \to A$ and we want to map this to a function $g: C \to A^B$ (which is a function which outputs functions).

So we need a map $h:A^{B \times C} \to (A^B)^C$ that inputs functions $f: B\times C \to A$ and whose output is a function $h(f): C \to A^B$. This means, we need to define the output function $h(f)$. The easiest way to do this, is to define $h(f)$ pointwise, but let's double check exactly what that means. The function $h(f)$ has domain $C$ and codomain $A^B$. This means if I take some $c \in C$ and plug it into $h(f)$ (i.e. we are looking at $h(f)(c)$) the output is some function $k: B \to A$. So for each $c \in C$, $h(f)(c)$ is a function which inputs $b \in B$ and outputs some $a \in A$. In the end, to completely define $h(f)$ I need to say how it will input any $c \in C$ and $b \in B$.

But this will end up working nicely, because for $f \in A^{B \times C}$ I know what $f$ does to the ordered pair $(b,c)$. So we let $h:A^{B \times C} \to (A^B)^C$ be defined pointwise as $$h(f)(c)(b) = f(b,c) \quad\text{ make sure you understand what this is doing!}$$

Now we must check that this $h$ is a bijection.

To see this is one-to-one suppose $f_1, f_2 \in A^{B\times C}$ were such that $h(f_1) = h(f_2)$. This means that $h(f_1)(c)(b) = h(f_2)(c)(b)$ for all $b \in B$, $c \in C$. But the definition of $h$ tells us then that $$ f_1(b,c) = h(f_1)(c)(b) = h(f_2)(c)(b) = f_2(b,c)$$ and so $f_1 = f_2$ and thus, $h$ is one-to-one.

Do you want to try and show that $h$ is onto?

Ryan Sullivant
  • 1,634
  • Thank you very much for your answer, now I understand better the concepts involved in this problem. To show that $h$ is onto, let $f(b,c) \in (A^B)^C$, so there exists $h(f):C→A^B$. This implies that there exists some $c \in C$ and some $b \in B$ such that $h(f)(c)(b)= f(b,c)$ (by the definition of $A^B$). Am I missing something? – Lstoi Apr 29 '14 at 03:53
  • I think you are going a little too quick there. It should be more like this: given some function $g \in (A^B)^C$ we need to find some function $f \in A^{B \times C}$ such that $h(f) =g$. So in this case we are given the function in the range, and we need to find the function in the domain that is mapped to it. In other words, we need to see that the function that would map to it is actually in the domain. Can you see what function would map to $g$? (Remember it is easier to look at how $g$ acts on it's points than just $g$ itself). Is that function an element of $A^{B \times C}$? – Ryan Sullivant Apr 29 '14 at 07:14
  • I'm stuck, I can't see what function would map to $g$. – Lstoi May 01 '14 at 15:58
  • Since $g \in (A^B)^C$ it is determined by its values on any $c \in C$ and $b \in B$. What about the function $f(b,c) = g(c)(b)$. Is $ f \in A^{B \times C}$? Where would $f$ get mapped (i.e what is $h(f)$)? – Ryan Sullivant May 01 '14 at 18:52
  • Yes, $f \in A^{B×C}$ and $f$ is mapped in $A$. – Lstoi May 01 '14 at 19:11