# Quick Question on Pre-image Terminology

Sorry for the daft question, but, is the following a correct thing to say?

"The preimage of a function f is a function iff for any element b in the range, there exists exactly one a in the domain such that f(a) = b"

"A function is injective iff its pre-image is a function."

Thank you.

Good question; terminology is confusing. It seems that you are using the term "preimage" incorrectly.

Suppose $f:A \to B$ is a function. For some set $D\subseteq B$, we say that $f^{-1}(D)$ is the preimage of $B$ under $f$.

The relation $f^{-1}: B \to A$ given by $f^{-1}(b) = \{a \in A: f(a) = b\}$ can be called the inverse (relation) of $f$. So, to restate what you meant correctly:

"The inverse relation of a function $f$ is a function iff for any $b$ in the (image), there exists exactly one $a$ such that $f(a) = b$"

"A function is injective iff its inverse relation is a function".

• Is there any particular reason why the set containment is strict? – Jonathan Hebert May 1 '14 at 0:14
• So are you saying that the preimage is strictly something one should consider to be a set, while the inverse relation is what one should consider as the "function" that is analogous to the preimage? – Jonathan Hebert May 1 '14 at 0:17
• Sorry, set containment is not necessarily strict. I generally use $\subsetneq$ when I mean strict containment, but I'll fix that. – Ben Grossmann May 1 '14 at 0:46
• And that's basically right. The preimage always refers to a set. The function that takes in a set and spits out its preimageis "the inverse relation", or "the inverse of the function", or even just "the inverse". – Ben Grossmann May 1 '14 at 0:49

You are misusing the term "preimage". I think you mean "inverse". If $f:X\rightarrow Y$ and $U\subset Y$, then the preimage of $U$ (under $f$) is the set $f^{-1}(U)=\{x\in X:f(x)\in U\}$.

If you replace the word "preimage" with the word "inverse", the your statements are correct. I presume that by "range" you mean "image" (in my example, this would be the set $f(X)\subset Y$, which is not necessarily all of $Y$ unless $f$ is also surjective).