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(Godement cours d'algèbre chapter two problem 5)

Let $X$, $Y$, $Z$ be three sets. Let $E$ be the set of all functions from $X\times Y$ to $Z$ and $F$ be the set of all functions from $X$ to $Z^Y$ (set of all functions from $Y$ to $Z$). Define a bijection between $E$ and $F$.

This is very confusing to me. Is the identity map of $X\times Y\times Z$ useful in figuring out a solution? I just want some pointer to how to look at it. I may have a solution (I can post it when I get home), but any help to understand the problem would be great.

Zev Chonoles gave a very helpful breakdown in the answer below. So I am focusing first on simply defining a function $E\to F$. What I have:

For any triplet $(a,b,f(a,b))\in X\times Y\times Z$, there is a function $h: Y\to Z$ such that $h(b)=f(a,b)$ and there is a function $g: X\to Z^Y$ such that $g(x)=h$. Then we can define the function $\varphi: E\to F$ such that $\varphi(f)=g$.

Is that correct?

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Hint: First, note that your goal is to come up with a function $\phi:E\to F$, i.e.., a rule for taking an element $f\in E$ and producing an element $\phi(f)\in F$. (You can worry later about proving that your function $\phi$ is a bijection.) That is, your goal is come up with a rule $\phi$ for taking a function $f:X\times Y\to Z$ and producing a function $\phi(f): X\to Z^Y$.

Now, a function $X\times Y\to Z$ has

  • input: $\fbox{$\text{element of }X$}$ $\fbox{$\text{element of }Y$}$
  • output: $\fbox{$\text{element of }Z$}$

and a function $X\to Z^Y$ has

  • input: $\fbox{$\text{element of }X$}$

  • output: a function with

    • input: $\fbox{$\text{element of }Y$}$
    • output: $\fbox{$\text{element of }Z$}$

Do you see how these ultimately amount to the same information (taking elements of $X$ and $Y$ as input, and producing elements of $Z$)? Can you formalize your observation as a definition for a function $\phi:E\to F$?

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    I see it is the same information. How about taking $(a,b,c)\in X\times Y\times Z$ and a function $f\in E$ such that $f(a,b)=c$. Since $Z^Y$ contains all the functions $Y\to Z$, it contains $g: b\mapsto c$. And since $F$ contains all the functions $X\to Z^Y$, it contains the function $a\mapsto g$. Does that get me close – 2017-02-16
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    I guess I hadn't finished above. I'll try to clarify also. For any triplet $(a,b,f(a,b))\in X\times Y\times Z$, there is a function $g: Y\to Z, b\mapsto f(a,b)$ and there is a function $h: X\to Z^Y, a\mapsto g$. Then we can define the function $\varphi: E\to F, f\mapsto h$. – 2017-02-17