Preimages of a Function

The preimage (or inverse image) of an element $ y \in Y $ in the codomain of a function $ f: X \to Y $, is the set of all elements $ x \in X $ in the domain that are mapped to $ y $ - in other words, all $ x $ such that $ f(x) = y $. The preimage of $ y $ is denoted by: $$ f^{-1}(\{y\}) = \{ x \in X \mid f(x) = y \} $$

Put simply, the preimage of an element $ y $ in the codomain $ Y $ consists of all those values in the domain $ X $ which, when passed through the function $ f(x) $, yield $ y $ as output.

domain and images

More generally, the preimage of a subset of $ Y $ under a function $ y = f(x) $ is the set of all domain elements $ X $ that the function maps into that subset of the codomain.

The preimage of a set tells us which elements in the domain map into a given subset via the function $ f $.

Note that the preimage $ f^{-1}(y) $ of an element $ y $ is not necessarily the inverse function, since it may associate one element in the codomain with multiple elements in the domain (that is, it may define a relation rather than a true function).

example of preimages of the same codomain element

This happens because a function is not required to be injective - a single codomain element may correspond to multiple domain elements.

preimages of y

The preimage is always well-defined, regardless of whether the function is injective or surjective.

Moreover, the preimage of a subset may be empty if no elements in the domain map into it.

Difference between a function and a relation. A function associates each element of the domain with exactly one element of the codomain. A relation, by contrast, may associate an element of the codomain with one or more elements of the domain. Thus, while every function is a relation, not every relation is a function.

A Practical Example
Properties of Preimages

A Practical Example

Consider the function $ f(x) = x^2 $.

The domain is the set of real numbers $ \mathbb{R} $, and the codomain is the set of non-negative real numbers $ \mathbb{R}^{\geq 0} $.

If we take an element in the codomain, say $ 4 $, then the preimage of $ 4 $ is the set of all $ x $ in the domain such that $ x^2 = 4 $:

$$ f^{-1}(\{4\}) = \{ -2, 2 \} $$

In this case, $ -2 $ and $ 2 $, since both $ (-2)^2 = 4 $ and $ 2^2 = 4 $.

preimages of 4 under the function y=x^2

Thus, the preimage of $ 4 $ is the set $ \{ -2, 2 \} $, as these are precisely the domain elements that map to $ 4 $ under $ x^2 $.

Note. In this example, the preimages in $ Y $ do not define an inverse function - rather, they define an inverse relation.

Properties of Preimages

Key properties of preimages under a function:

Preimage of a union
The preimage of the union of two sets equals the union of the preimages: $$ f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B) $$
Preimage of an intersection
The preimage of the intersection of two sets equals the intersection of their preimages: $$ f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B) $$
Preimage of a complement
The preimage of the complement of a set equals the complement of the preimage: $$ f^{-1}(Y \setminus A) = X \setminus f^{-1}(A) $$

These are sometimes referred to as the preimage preservation properties under union, intersection, and difference.

Example

Now consider the function:

$$ f(x) = \sqrt{x} $$

where $ f: \mathbb{R}^{\geq 0} \to \mathbb{R}^{\geq 0} $, since the square root is only defined for $ x \geq 0 $.

Let us choose two subsets of the domain:

$$ A_1 = [0, 1) $$

$$ A_2 = [4, 9] $$

The preimage of the interval $ A_1 = [0, 1) $ is:

$$ f^{-1}(A_1) = \{ x \in \mathbb{R}^{\geq 0} \mid \sqrt{x} \in [0, 1) \} = [0, 1) $$

The preimage of $ A_2 = [4, 9] $ is:

$$ f^{-1}(A_2) = \{ x \in \mathbb{R}^{\geq 0} \mid \sqrt{x} \in [4, 9] \} = [16, 81] $$

Now let us check the three set operations:

A] Union

The union of the preimages is:

$$ f^{-1}(A_1) \cup f^{-1}(A_2) = [0, 1) \cup [16, 81] $$

The union of the intervals is:

$$ A_1 \cup A_2 = [0, 1) \cup [4, 9] $$

Thus, the preimage of the union is:

$$ f^{-1}(A_1 \cup A_2) = f^{-1}([0, 1) \cup [4, 9]) = [0, 1) \cup [16, 81] $$

This confirms that $ f(x) = \sqrt{x} $ preserves unions:

$$ f^{-1}(A_1 \cup A_2) = f^{-1}(A_1) \cup f^{-1}(A_2) $$

B] Intersection

The intersection of the preimages is the empty set, since there is no overlap between $ [0, 1) $ and $ [4, 9] $:

$$ f^{-1}(A_1) \cap f^{-1}(A_2) = [0, 1) \cap [16, 81] = \emptyset $$

Likewise, the preimage of the intersection is also empty:

$$ f^{-1}(A_1 \cap A_2) = f^{-1}(\emptyset) = \emptyset $$

Again, this confirms that $ f(x) = \sqrt{x} $ preserves intersections:

$$ f^{-1}(A_1 \cap A_2) = f^{-1}(A_1) \cap f^{-1}(A_2) $$

C] Complement

Now consider the complement of $ A_1 = [0, 1) $ in $ \mathbb{R}^{+} $:

$$ \mathbb{R}^{+} \setminus A_1 = [1, \infty) $$

The preimage of the complement is:

$$ f^{-1}(\mathbb{R}^{+} \setminus A_1) = f^{-1}([1, \infty)) = [1, \infty) $$

The complement of the preimage of $ A_1 $ is:

$$ \mathbb{R}^{+} \setminus f^{-1}(A_1) = \mathbb{R}^{+} \setminus [0, 1) = [1, \infty) $$

Thus, $ f(x) = \sqrt{x} $ also preserves complements:

$$ f^{-1}(\mathbb{R}^{+} \setminus A_1) = \mathbb{R}^{+} \setminus f^{-1}(A_1) $$

In this example, all the key preimage properties with respect to union, intersection, and complement are satisfied.

And so on.