Functional notation usage for arbitrary relation? Question about questionable notation used in a homework assignment.
The assignment defines a relation f := { (0,1), (1,3), (2,1) }. It then asks to "show" this is a function. IMO strictly (i.e. Bourbaki) speaking it is not but so be it. The intention is clear.
It further asks to calculate f[2], f(2), f-1[1], f-1(1), ...
It is understood that the author uses Kuratowski's definition of ordered pairs, the natural numbers as n = {0, ..., n-1} and that f-1 is the inverse relation
f-1 = { (1,0), (3,1), (1,2) }, which is obviously not a function.
So my question is: what could f-1(.) possibly mean for the non-functional relation f-1? It is sometimes used as a sloppy alternative notation for f-1[.] but it is obvious from the question that the author makes a strict distinction between [.] and (.). And that whatever (.) is supposed to mean, its definition should generalize Euler's notation (.) for functions.
 A: We have the well-known set-theoretic definition of relation (a set of pairs, for a binary one) and the definition of function, i.e. a relation which is "functional".
For a relation $R$ we define its converse :

for a binary relation is the relation that occurs when the order of the elements is switched in the relation.

The converse relation of a function always exist; if it is a function, we say that the function is invertible, in which case the converse relation is the inverse function.
The symbol for the converse of $R$ is not very standard; we have $\breve{R}$, but sometimes $R^{-1}$ is used.
Having said that, for $f = \{ (0,1), (1,3), (2,1) \}$, we have that $f^{-1} =  \{ (1,0), (3,1), (1,2) \}$, in which case $f^{-1}(1) = \{ 0, 2 \}$.

Added after the comments exchange.
Ref. to Herbert Enderton, Elements of set theory (Academic Press, 1977), page 46.
Definitions :

The inverse [most commonly applied to functions, sometimes are applied to relations] of $F$ is the set $F^{-1} = \{ (u,v) \mid (v,u) \in F \}$
The restriction of $F$ to $A$ is the set $F \restriction A = \{ (u,v) \mid (u,v) \in F \text { and } u \in A \}$.
The image of $A$ under $F$ is the set $F[A] = \text {Range}(F \restriction A) = \{ v \mid (\exists u \in A) ((u,v) \in F) \}$.

An interesting example follows :

Let $F = \{ (\emptyset, a), ( \{ \emptyset \}, b) \}$. Observe that $F$ is a function. We have $F^{-1} = \{ (a, \emptyset), (b, \{ \emptyset \}) \}$ which is a function iff $a \ne b$.
The restriction of $F$ to $\emptyset$ is $\emptyset$, but $F \restriction \{ \emptyset \} = \{ ( \emptyset , a) \}$.

Consequently, $F[ \{ \emptyset \} ] = \{ a \}$, in contrast to the fact that $F( \{ \emptyset \})= b$.


Now we can try to answer the problem above, considering your statement that for the author : "the natural numbers are $n = \{ 0, \ldots, n-1 \}$."
We have that $f(2)=1$ (an element) while $f[2]=f( \{ 0,1 \} ) = \{ 1,3 \}$ (a set).
In the same way, if we have to be consistent with the notation, we must have :

$f^{-1}[1]=f^{-1}[ \{ 0 \}]= \emptyset$, because there are no pairs $(0,x)$ in $f^{-1}$,

while $f^{-1}(1)= \{ 0,2 \}$, because $f^{-1}$ is not a function.
A: The inverse of a function f is f$^{-1}$ when the inverse of f exists.
The inverse of a relation R is R$^{-1}$ = { (x,y) : yRx }.
The set extensions of a function f are f[A] = { f(x) : x in A }
and f$^{-1}$[A] = { x : f(x) in A }.  
