I am trying to step through a proof provided by my book of this statement:
$^C(^B A) \approx \ ^{C \times B}A$
The author writes the following:
Define $\Phi : \ ^C(^B A) \xrightarrow[onto]{1-1}\ ^{C \times B}A $ by saying that $\Phi (f)(c,b) = \big(f(c)\big)(b)$
I have never seen these notations before: $\Phi (f)(c,b)$ or $\big(f(c)\big)(b)$. What do they signify?
$^C(^B A)$ and $^{C \times B}A$ are both sets of functions...but I am not certain how to parse the above notation.
Focusing just on $\ ^C(^B A) $, I can decompose this set into first $^B A$, which is a set of functions from $B$ to $A$. Which I'll illustrate as $\{g_{B \to A_1}, g_{B \to A_2},...,g_{B \to A_n}\}$.
Then $\ ^C(^B A) $ is the set of functions from $C$ to $\{g_{B \to A_1}, g_{B \to A_2},...,g_{B \to A_n}\}$.
So one such function $f$ would look like: $f_1= \{\langle c_1, g_{B \to A_1} \rangle,...,\langle c_m,g_{B \to A_m} \rangle \}$
Given that $g_{B \to A_1}$ would look something like $\{\langle b_1, a_1 \rangle, ..., \langle b_p, a_p \rangle \}$, the first element of $f_1$...i.e. $\langle c_1, g_{B \to A_1} \rangle$...would look like:
$\langle c_1, \{\langle b_1, a_1 \rangle, ..., \langle b_p, a_p \rangle \} \rangle$
Thus, $f_1$ contains different variations of this.
But even after breaking this down, I still do not see what $\Phi (f)(c,b)$ is specifically referencing.
Since the domain of $\Phi$ is ${^C}\left({^B}A\right)$, the $f$ in $\Phi(f)$ must be a function from $C$ to ${^B}A$. The codomain of $\Phi$ is ${^{C\times B}}A$, so $\Phi(f)$ must be a function from $C\times B$ to $A$; for convenience call this function $g$, so that $\Phi(f)=g:C\times B\to A$. Now we can apply this function $g$ to any $\langle c,b\rangle\in C\times B$ to get $g(c,b)\in A$. And $g$ is just a simpler name for the function $\Phi(f)$, so we can equally well write this element of $A$ as $\big(\Phi(f)\big)(c,b)$. Finally, context makes it clear that $\Phi(f)$ is a function from $C\times B$ to $A$, so we can drop one set of parentheses and write simply $\Phi(f)(c,b)$. That is, we know that the function $\Phi$ takes a single argument, so $f$ must be that argument, and we know that $\Phi(f)$ is a function that takes arguments in $C\times B$, so $\langle c,b\rangle$ must be the input to the function $\Phi(f)$.
The rest of the quoted definition tells you which function from $C\times B$ to $A$ the function $\Phi(f)$ actually is. Recall that $f$, the input to $\Phi$, must be a function from $C$ to ${^B}A$. And $c\in C$, so we can apply $f$ to $c$ to get a function $f(c)$ from $B$ to $A$. For convenience I’ll call that function $h$, so that $f(c)=h:B\to A$. Now $b\in B$, the domain of $h$, so $h(b)$ makes sense and is some element of $A$, say $a$. Finally, we can get rid of the abbreviation $h$: $h$ is just handier name for the function $f(c)$, so $h(b)=a$ is just an abbreviation for $\big(f(c)\big)(b)=a$.
To sum up, for each $f:C\to{^B}A$ we define a function $\Phi(f):C\times B\to A$ by letting $\Phi(f)$ take the ordered pair $\langle c,b\rangle\in C\times B$ to the same element of $A$ to which the function $f(c):B\to A$ takes $b$.