Krdytkyu

Show that if $A approx B$ then $A^C approx B^C$

According to my textbook, $A^C$ is the set of all functions from $C$ to $A$, ie. $A^C={Fin mathscr P(Y times X): F$ is a function$}$.

So I need to prove that there is a function $g: A^C to B^C$ that is one-one and onto. ie. For one-one, assume $g(h)=g(h')$ where $h$ is a function in $A^C$, and show that $h=h'$. Also assume that there is a $f: Ato B$ that is one-one and onto.

I am stuck because I have no idea how to define $g$; I think this is partly because I am slightly confused by what the definition of $g: A^C to B^C$ is. I know it is a set of functions, but what exactly does that mean/ what does a member look like? Is it a set of ordered pairs consisting of functions? eg. $<f,g>$?

Also it seems that some info about $C$'s relation to other sets is needed, but is not provided.

Could anyone help or point me at the right direction please?

(Schröder–Bernstein theorem is available, but I am not sure if it will be useful here)

Technically functions can be modeled by certain sets of pairs, but that is not a very productive intuition about what is going on for your purpose here.

I think it is more fruitful to think of a function $Xto Y$ as a machine with the property that when you put an element of $X$ into it, somehow an element of $Y$ comes out the other end -- and this happens in a repeatable fashion such that if you put the same element of $X$ into the machine multiple times, the same element of $Y$ will be produced several times.

When you're looking for a function $g:A^C to B^C$, then you want to constructs a machine that makes machines into other machines. The input to your $g$ will be a machine $Cto A$. You want use that to produce a machine $Cto B$.

The natural way to do that will be to bolt the $Cto A$ machine you get as input together with a copy of the the $f:Ato B$ you have assumed exists. The combined machine will first use the input to convert the incoming $c$ into an $a$, and then use $f$ to convert the $a$ into a $b$. The net effect is that something from $C$ comes in and something from $B$ comes out -- in other words we do get a function $Cto B$.

Once you have used the machine analogy to work out what your $g$ should do, you can go back to symbolic reasoning to try to prove that this $g$ is actually a bijection.

Given an element of $A^C,$ it is some map $u:C to A.$ Then define $g(u):C to B$ by composition. That is, $g(u)(c)=f(u(c)).$

Since A$approx$B, there exists a one-to-one onto function f:A$rightarrow$B. That is the key to the exercise.

Let $phiin A^C$. Then f$circ phi$:C$rightarrow$B. So we define a function $psi_f$:$A^C rightarrow B^C$ such that $phi mapsto fcircphi$.

1:$psi_f$ is onto. Let g$in B^C$. Then $f^{-1}circ g in A^C$. So
$$psi_f(f^{-1} circ g) = f circ f^{-1} circ g = g. checkmark $$

2.$psi_f$ is one-to-one. Suppose $phi_1,phi_2 in A^C$ such that $f circ psi_1 = f circ psi_2$. Thus for every c $in$ C, then
$$f(psi_1(c)) = f(psi_2(c))$$
But since f is invertible, that forces
$$psi_1(c) = psi_2(c)text{ } forall c checkmark$$