Notation for set of functions between two sets

Question

I have recently come across the notation $\mathcal{B}^{\mathcal{A}}$ meaning the set of all functions from set $\mathcal{A}$ to set $\mathcal{B}$.

Using this fact, I am wondering if the following expressions are equivalent: \begin{align*} &f:\mathcal{A}\times\mathcal{B}\to \mathcal{C}\\ &f:\mathcal{A}\to\mathcal{C}^\mathcal{B} \end{align*}

Apologies if this is trivial, I am just a bit confused about the usage of the set raised to another set notation.

Answer 1

No. Strictly speaking the two notations talk about two different functions simply because they have different domains and codomains:

• $f : A \times B \to C$ is a function that produces a value $f(a, b)$ in $C$ when applied to a pair $(a, b)$ in $A \times B$ whereas
• $f : A \to C^B$ is a function that itself produces a function $f(a) : B \to C$ when applied to a value $a$ in $A$.

However, they are equivalent in the following sense:

Given any function $f : A \times B \to C$, you can get a function $f^\ast : A \to C^B$ by the recipe: $$\big[f^\ast (a)\big](b) = f(a, b)$$ Conversely, given any $g : A \to C^B$, you can get a function $\overline{g} : A \times B \to C$ by the recipe: $$\overline{g}(a, b) = \big[g(a)\big](b)$$ The correspondences $f \mapsto f^\ast$ and $g \mapsto \overline{g}$ between the sets $A \times B \to C$ and $A \to C^B$ are inverses of one another and they establish a natural bijection between those two sets. In this sense you can see the two functions $f : A \times B \to C$ and $f^\ast : A \to C^B$ (or $g : A \to C^B$ and $\overline{g} : A \times B \to C$) as being essentially the same.

Some go as far as to drop the asterisk and say that $f : A \times B \to C$ and $f : A \to C^B$ are the same function when what they really mean is that $f : A \times B \to C$ and $f^\ast : A \to C^B$, as defined as above, represent the same function. This abuse of notation is probably what got you confused.