Image and Preimage
Image and preimage are connected and associated like individual elements are
Image and preimage are sets of elements (not of ordered pairs)
Image is a subset of codomain but defined by a subset of domain's elements
Preimage is a subset of domain but defined by a subset of codomain's elements
Definition: let f be a function f : A -> B. Then
If
X ⊆ A, the image ofXis the setf(X) = { f(x) : x ∈ X } ⊆ BIf
Y ⊆ B, the preimage ofYis the setf⁻¹(Y) = { x ∈ A : f(x) ∈ Y } ⊆ A
The image f(X) of X is the set of all things in B, where X ⊆ A, that f sends elements of X to. Roughly, f(X) may be considered a distorted "copy" (image) of X in the codomain B.
The preimage f⁻¹(Y) of Y, where Y is a subset of codomain B i.e. Y ⊆ B, is the set of all things in the domain A that f sends into Y.
If Y ⊆ B, the expression f⁻¹(Y) has a meaning even if f is not an invertible function; it denotes the set of elements in A, that is, it denotes all the elements in X, where X is a subset of the domain A.
Some authors talk about a single element x that is associated to the element f(x), which they call x's image, even though it is a single element rather than a singleton set (sloppy treatment). Also they call x the preimage of f(x). In a strict treatment image and pre-image are always sets, even if they contain only a single element. The sloppy treatment may be tolerated when a bijective function is scrutinize, but in the case of a surjective function it can easily happen that the image of an element x, {x}, is a set of more than one element, like {a,b}. In a bijective function, the associated image and preimage are always equinumerous sets.
Example
Let f : A -> B
where A = {s,t,u,v,w,x,y,z}
and B = {0,1,2,3,4,5,6,7,8,9}
and the function f is defined by the following set of ordered pairs:
f={ (s,4) , (t,8) , (u,8) , (v,1) , (w,2) , (x,4) , (y,6) , (z,4) }
The function f is neither injective nor surjective, so it certainly is not invertible.
It is important to realize that the preimage X, where X ⊆ A, is a subset of domain A, and the image Y, where Y ⊆ B, is a subset of codomain B. They are not individual elements nor sets of ordered pairs (like functions and relations are).
Still considering the example, if X ⊆ A and X = {s,x,z} and Y ⊆ B and B = {4}, then X is the preimage f⁻¹({s,z,x}) = {4} and Y is the image f({4}) = {s,z,x}. The expression f⁻¹(4) has no meaning since f has no inverse (function). Likewise, there is a difference between f({s}) = {s} and f(s) = 4.
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