mathematics-physics-wiki/docs/en/mathematics/set-theory/maps.md

Maps

Definition

Definition: a relation f from a set A to a set B is called a map or function from A to B if for each a \in A there is one and only one b \in B with afb.

• To indicate that f is a map from A to B we may write f:A \to B.
• If a \in A and b \in B is the unique element with afb then we may write b=f(a).
• The set of all maps from A to B is denoted by B^A.
• A partial map f from a A to B with the property that for each a \in A there is at most one b \in B with afb.

For example, let f: \mathbb{R} \to \mathbb{R} with f(x) = \sqrt{x} for all x \in \mathbb{R} is a partial map, since not all of \mathbb{R} is mapped.

Proposition: let f: A \to B and g: B \to C be maps, then the composition g after f: g \circ f = f;g is a map from A to C.

??? note "Proof:"

Let a \in A then g(f(a)) is an element in C in relation f;g with a. If c \in C is an element in C that is in relation f;g with a, then there is a b \in B with afb and bgc. But then, as f is a map, b=f(a) and as g is a map c=g(b). Hence c=g(b)=g(f(a)) is the unique element in C which is in relation g \circ f with a.

Definition: Let f: A \to B be a map.

• The set A is called the domain of f and the set B the codomain.
• If a \in A then the element b=f(a) is called the image of a under f.
• The subset of B consisting of the images of the elements of A under f is called the image or range of f and is denoted by \text{Im}(f).
• If a \in A amd b=f(a) then the element a is called a pre-image of b. The set of all pre-images of b is denoted by f^{-1}(b).

Notice that b can have more than one pre-image. Indeed if f: \mathbb{R} \to \mathbb{R} is given by f(x) = x^2 for all x \in \mathbb{R}, then both -2 and 2 are pre-images of 4.

If A' is a subset of A then the image of A' under f is the set f(A') = \{f(a) \;|\; a \in A'\}, so \text{Im}(f) = f(A).

If B' is a subet of B then the pre-image of B', denoted by $f^{-1}(B') is the set of elements a from A that are mapped to an element b of B'.

Theorem: let f: A \to B be a map.

• If A' \subseteq A, then f^{-1}(f(A')) \supseteq A'.
• If B' \subseteq B, then f(f^{-1}(B')) \subseteq B'.

??? note "Proof:"

Let a' \in A', then f(a') \in f(A') and hence a' \in f^{-1}(f(A')). Thus A' \subseteq f^{-1}(f(A')).

Let a \in f^{-1}(B'), then f(a) \in B'. Thus f(f^{-1}(B')) \subseteq B'.