bijection between two sets of functions

**issacnewton** · January 1st 2012, 06:34 AM

Hi

here's a problem I am trying to solve. Suppose A,B,C,D are sets

$[(A\sim B)\wedge(C\sim D)\Rightarrow (^AC \sim \; ^BD)]$

where $^AC$ is set of all functions from A to C. and $^BD$ is
set of all functions from B to D. Following is my work. I have already prove that

$[((A\sim B)\wedge(C\sim D)\Rightarrow(A\times C \sim B \times D)]$

Suppose

$(A\sim B)\wedge(C\sim D)$

$\therefore A\times C \sim B \times D$

I have also already proved that, for any sets A and B,

$(A\sim B)\Rightarrow(\mathcal{P}(A)\sim \mathcal{P}(B))$

$\therefore \mathcal{P}(A\times C) \sim \mathcal{P}(B \times D)$

that means that there is a bijection from $\mathcal{P}(A\times C)$
to $\mathcal{P}(B \times D)$ .

$\mathcal{P}(A\times C)$ is the set of all subsets of $A\times C$
. But subsets of $A\times C$ are just relations from A to C. So
$\mathcal{P}(A\times C)$ is set of all relations from A to C. In similar way,
$\mathcal{P}(B\times D)$ is set of all relations from B to D. Functions are special kind of relations , So

$^AC\subseteq \mathcal{P}(A\times C)$

$^BD \subseteq \mathcal{P}(B\times D)$

Since $\mathcal{P}(A\times C) \sim \mathcal{P}(B\times D)$ , there is some bijection between these two sets. Let it be h

$h:\mathcal{P}(A\times C)\to \mathcal{P}(B\times D)$

From here on, I am stuck. We have a bijection between one family of relations to another family of relations. What we need is a bijection between the subsets of
these families.

**FernandoRevilla** · January 1st 2012, 09:29 AM

Originally Posted by issacnewton

Suppose A,B,C,D are sets

$[(A\sim B)\wedge(C\sim D)\Rightarrow (^AC \sim \; ^BD)]$

where $^AC$ is set of all functions from A to C. and $^BD$ is
set of all functions from B to D.

I suppose you mean functions or mappings. Then, consider bijections $\pi:A\to B$ and $\phi: C\to D$ . Now, define $\Gamma:^AC\to ^BD$ such that $\Gamma (f)=\phi\circ f\circ \pi^{-1}$ . Easily proved, $\Gamma$ is bijective hence $^AC \sim \; ^BD$ .

Edited: Wrong argument.
Edited: Wrong day for posting.

**emakarov** · January 1st 2012, 09:45 AM

Here is a picture.

Note that since $\pi$ and $\phi$ are bijections, there exist $\pi^{-1}$ and $\phi^{-1}$ such that $\pi\circ\pi^{-1}=\mathrm{id}_B$ , $\pi^{-1}\circ\pi=\mathrm{id}_A$ , $\phi\circ\phi^{-1}=\mathrm{id}_D$ and $\phi^{-1}\circ\phi=\mathrm{id}_C$ , where $\mathrm{id}_X$ is the identity function on $X$ . Therefore, you can multiply various equalities by these functions. E.g., $\Gamma (f)=\phi\circ f\circ \pi^{-1}$ implies $\Gamma (f)\circ\pi= \phi\circ f\circ \pi^{-1}\circ\pi =\phi\circ f\circ \mathrm{id}_A= \phi\circ f$ .

**FernandoRevilla** · January 1st 2012, 10:03 AM

Right. A quick, mental (and wrong) reasoning led me to $\Gamma (f)=\Gamma (g)\Rightarrow\ldots \Rightarrow f=g$ . Better appproach: denoting $a=\textrm{card} (A),\ldots, \textrm{card}(D)=d$ we have $a=b,\;c=d$ , $\textrm{card}(^AC)=c^a$ etc.

**emakarov** · January 1st 2012, 10:17 AM

Um, why is it wrong?

**FernandoRevilla** · January 1st 2012, 10:30 AM

Originally Posted by emakarov

Um, why is it wrong?

Perhaps too much whisky and champagne last night.

**issacnewton** · January 2nd 2012, 01:18 AM

thanks. I was almost there as I have proved the bijection between the set of relations from A to C to the set of relations from B to D.

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