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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.

Actually this question hasn't been posted by me, but has already been answered and closed as Find a bijection from $(A^B)^C$ into $A^{B \times C}$

I don't agree, since this doesn't seen at least for me to be correct. Maybe I haven't got through the answer but in my view, the correct answer should be, following the same letters for the functions:

my Answer

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

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

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

I would like to know if my editions to the functions really answer the question or if the previous answer Find a bijection from $(A^B)^C$ into $A^{B \times C}$ was indeed correct. Thanks.

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    You wrote *it should be written$f(c)(b)$instead of$f(b)(c)$shouldnt it?* But in [the answer](http://math.stackexchange.com/a/57399/) you are talking about we have this: $\varphi(f)(b,c)=f(c)(b)$. Exactly as you suggest. (Or perhaps I've misunderstood what exactly the problem is.)2012-10-19

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Your argument looks like this:

Show that there is a bijection between set $X$ and set $Y$.

Let $x\in X$, $y\in Y$. Define $\Phi\colon X\to Y$ by setting $\Phi(x)=y.$ This is a bijection because it has an inverse $\Psi\colon Y\to X$ $\Psi(y)=x.$

Do you see that this is not ok?

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    @user45147 I am not sure what you mean in your comment. *Who* do you mean, when you say "he" in your comment. *Where* do you want to change $f(b)(c)$ to $f(c)(b)$.2012-10-19