# Function Images

# Why

We consider the set of results of a set of
domain elements.

# Definition

The image of a set of
domain elements under a function is the set of
their results.
Though the set of domain elements may include
several distinct elements, the image may still
be a singleton, since the function may map all
of elements to the same result.

Using this language, the range (see Functions) of a function is the image of its domain.
The range includes all possible results of the
function.
If the range does not include some element of
the codomain, then the function maps no domain
elements to that codomain element.

## Notation

Let $f: A \to B$.
We denote the image of $C \subset A$ by
$f(C)$, read aloud as “f of C.”
This notation is overloaded: for every $c \in
C$, $f(c) \in A$, whereas $f(C) \subset A$.
Read aloud, the two are indistinguishable, so
we must be careful to specify whether we mean
an element $c$ or a set $C$.
Following this notation for function images, we
denote the range of $f$ by $f(A)$.
In this notation, we can record that $f$ maps
$X$ onto $Y$ by $f(X) = Y$.

## Notational ambiguity

The notation $f(A)$ is can be ambiguous in the
case that $A$ is both an element and a set of
elements of the domain of $f$.
For example, consider $f: \set{\set{a}, \set{b},
\set{a, b}} \to X$.
Then $f(\set{a, b})$ is ambiguous.
We will avoid this ambiguity by making clear
which we mean in particular cases.

# Inverse images

Similarly to how we can define $f: \powerset{X}
\to \powerset{Y}$ for $A \subset X$

\[
f(A) = \Set{y \in Y}{(\exists x)(x \in a \land y = f(x))},
\]

we can define $f^{-1}: \powerset{Y} \to
\powerset{X}$ for $B \subset X$

\[
f^{-1}(B) = \Set{x \in X}{(\exists y)(y \in B \land y =
f(x))}.
\]

In other words, $f^{-1}(B)$ is the set of all
elements of the domain which give the elements
in $B$ of the range.
We call $f^{-1}(B)$ the inverse
image of $B$.
Another name less commonly used is
counter image or
counterimage.
# Connections

Here are some connections.
Let $f: X \to Y$ and $B \subset Y$.
$f(f^{-1}(B)) \subset B$. If $f$ is onto, then
$f(f^{-1}(B)) \subset B$.

Let $f: X \to Y$ and $A \subset X$.
$A \subset f^{-1}(f(A))$.
If $f$ is one-to-one, then $A = f^{-1}(f(A))$.