Is it acceptable to use the inverse notation for certain elements of a non-bijective function? Say we have the function $f:A \rightarrow B$ which is pictured below.
This function is not bijective, so the inverse function $f^{-1}: B \rightarrow A$ does not exist. However, we can see that elements $d$ and $e$ in set $B$ are each mapped to by a single element in $A$, so they are kind of "nice" in this sense, almost like a bijective function.
Would it be acceptable to use the notation $f^{-1}(d)=6$ and $f^{-1}(e)=7$ here, even though not all the elements of $B$ have a single inverse? Or does the use of $f^{-1}$ always imply that the entire function $f$ has an inverse?
 A: No, but it does make sense to say that $f^{-1} (\{d\}) = \{6\}$ (which reads as: the preimage of the set $\{d\}$ is $\{6\}$). This is standard mathematical notation. This also allows you to say that $f^{-1} (\{c\}) = \emptyset$, so it's a very convenient notation.
Note that with this notation, you must now view the formula $f^{-1} (B)$ as an indecomposable formula, i.e. you are not talking about the function $f^{-1}$ applied to some subset $B$, because $f^{-1}$ does not exist.
A: If it is not injective then no, because we don't know what preimage you are referring to when you write $f^{-1}(x)$ if $x$ has multiple preimages. But what you can write is $f^{pre}(x)$ indicating the preimage  (or set of preimages) of $x$. 
A: Yes that's perfectly reasonable and it's done with regularity.  BUT you have to interpret $f^{-1}(x)$ as a set, a subset of the domain, and not as an element of the domain.  That's the case even if the set has one element.  Just make sure you don't treat $f^{-1}$ as a real function and the notation is fine.
