# Show that there exists a bijection from $(A^B)^C$ into $A^{B \times C}$

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:

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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If you want to define $\Phi: (A^B)^C \to A^{B \times C}$, then $\Phi(f)$ should be an element of $A^{B \times C}$, i.e., a function from $B\times C$ to $A$. So you plug pairs (elements of $B\times C$) into the function $\Phi(f)$. –  Martin Sleziak Oct 18 '12 at 19:02
I am very confused as for why this question has a duplicate banner. (And generally why is it a copy-paste of math.stackexchange.com/questions/178277 for its first half) –  Asaf Karagila Oct 18 '12 at 19:04
@user45147 I believe that it would be better if, instead of putting the identical text as the stackexchange software includes in questions which are closed, you would explain that your question is related to the other one and that are in fact asking about clarification of one point of the proof. Otherwise it looks very confusing (as Asaf mentioned in his comment). We are all used to see that banner on closed questions only. –  Martin Sleziak Oct 18 '12 at 19:16
ok... i am going to clarify the answerc, and sorry, what banner do you talk about martin ? I'm new to this –  user45147 Oct 18 '12 at 20:51
Ok @MartinSleziak but then i think that it should be written $f(c)(b)$ instead of $f(b)(c)$ shouldnt it ? –  user45147 Oct 18 '12 at 22:31

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 $\Psi\colon Y\to X$ $$\Psi(y)=x.$$
See Bijection iff Left and Right Inverse at ProofWiki. If we show that $\Phi$ has an inverse, then $\Phi$ is bijective. This is what is done in the answer to linked question. OP is asking whether definition of $\Phi$ and $\Psi$ suggested there works. (He proposes another maps.) Of course, after we define $\Phi$ and $\Psi$, we must also show that they are inverse to each other. I certainly agree with that. –  Martin Sleziak Oct 19 '12 at 6:23
so @MartinSleziak you are saying that the proof in the answer is incomplete, thats it ? Although he has defined a function from, say, x to y, and another from y to x, he has not shown that they are inverse to each other, what is left to be proved on the exercise. is that ? And about that I said above, that it should be $f(c)(b)$ instead of $f(b)(c)$ ? Thanks for your answer –  user45147 Oct 19 '12 at 16:03
@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 $f(c)(b)$. –  Martin Sleziak Oct 19 '12 at 17:32