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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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    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 $\Phi(f)$.2012-10-18
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    I am **very** confused as for why this question has a duplicate banner. (And generally why is it a copy-paste of http://math.stackexchange.com/questions/178277 for its first half)2012-10-18
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    @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.2012-10-18
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    ok... i am going to clarify the answerc, and sorry, what banner do you talk about martin ? I'm new to this2012-10-18
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    Ok @MartinSleziak but then i think that it should be written $f(c)(b)$ instead of $f(b)(c)$ shouldnt it ?2012-10-18
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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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