# Injective maps from $B^A$ to $(C^B)^{C^A}$

Let $A$, $B$, and $C$ be nonempty sets. I need to find conditions to guarantee that there is an injective function from $B^A$, the set of all functions from $A$ to $B$, to $\displaystyle (C^B)^{C^A}$.

Afterwards, I need to show this kind of a function and to prove that it is injective one.

I tried to do that with a power equation. Any suggestions? Thanks (and if you may try to write it in simple as you can due to my math-English barriers :) )

-Nir

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You mean if $f: A \to B$ and $g: B \to C$ are injective, then $h = g(f(x))$ is injective from $A$ to $C$? –  Joshua Shane Liberman Feb 10 '11 at 18:25
No, actually the question is more as In what conditions will you have an injective functions from (A-->B) (the group of functions from Set A to set B) to the function between ((B-->C) to (A-->C)) –  user6163 Feb 10 '11 at 18:28
The usual set theoretic notation for the set of functions from A to B is $B^A$, so you're trying to show that there exists an injective map $\eta: B^A \to (C^B)^{(C^A)}$ –  kahen Feb 16 '11 at 3:04

First, define your mapping from $B^A$ to $(C^B)^{C^A}$: we're essentially forced in our definition. Namely, given $f:A\to B$ and $g:B\to C$ we need to produce a function $A\to C$ but this clearly must be given by $g \circ f$. (That's the only way to get domain and co-domain to match up correctly.) In other words, we define $\eta:B^A\to (C^B)^{C^A}$ by $\eta(f)(g)(x)=g(f(x))$.
Next we need to show that this map is injective. Suppose that $f_1,f_2:A\to B$ with $f_1\neq f_2$, then we need to show that $\eta(f_1)\neq\eta(f_2)$. Consider some point $a\in A$ with $f_1(a)\neq f_2(a)$ and go from there (basically just push through the definitions). (Remaining details left as an exercise as well as arguing why there is such an $a$.)
Let $C$ be a singleton. Then $C^A$, $C^B$, $C^A$, and hence $(C^B)^{C^A}$ are singletons. Take $A$ and $B$ such that $\mathrm{card}(A\to B)\gt 1$, e.g. $A=B=\{0,1\}$. (“card” is a number of elements in a set, cardinality.) Then $\mathrm{card}(B^A) \gt \mathrm{card}((C^B)^{C^A})$, so there is no injection.