luc/notes
1
0
Fork 0
Code Issues Pull requests Activity
notes/docs/mathematics/set-theory/maps.md

5.8 KiB
Raw Blame History

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 $f(f^{-1}(B')) \subseteq B'$.

Special maps

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

  • f is called surjective, if for each b \in B there is at least one a \in A with b = f(a). Thus \text{Im}(f) = B.
  • f is called injective if for each b \in B, there is at most one a with f(a) = b.
  • f is called bijective if it is both surjective and injective. So, if for each b \in B there is a unique a \in A with f(a) = b.

For example the map \sin: \mathbb{R} \to \mathbb{R} is not surjective nor injective. The map \sin: [-\frac{\pi}{2},\frac{\pi}{2}] \to \mathbb{R} is injective but not surjective and the map \sin: \mathbb{R} \to [-1,1] is surjective but not injective. To conclude the map \sin: [-\frac{\pi}{2},\frac{\pi}{2}] \to [-1,1] is a bijective map.


Theorem: let A be a set of size n and B a set of size m. Let f: A \to B be a map between the sets A and B.

  • If n < m then f can not be surjective.
  • If n > m then f can not be injective.
  • If n = m then f is injective if and only if it is surjective.

??? note "Proof:"

Think of pigeonholes. (Not really a proof).

Proposition: let f: A \to B be a bijection. Then for all a \in A and b \in B we have f^{-1}(f(a)) = a and f(f^{-1}(b)) = b. In particular, f^{-1} is the inverse of f.

??? note "Proof:"

Let $a \in A$. Then $f^{-1}(f(a)) = a$ by definition of $f^{-1}$. If $b \in B$ then by surjectivity of $f$ there is an $a \in A$ with $b = f(a)$. So, by the above $f(f^{-1}(b)) = f(f^{-1}(f(a))) = f(a) = b$.

Theorem: let f: A \to B and g: B \to C be two maps.

  1. If f and g are surjective then so is g \circ f.
  2. If f and g are injective then so is g \circ f.
  3. If f and g are bijective then so is g \circ f.

??? note "Proof:"

1. Suppose $f$ and $g$ are surjective, let $c \in C$. By surjectivity of $g$ there is a $b \in B$ with $g(b) = c$. Since $f$ is surjective there is also an $a \in A$ with $f(a) = b$. Therefore $g \circ f(a) = g(f(a)) = g(b) = c$.
2. Suppose $f$ and $g$ are injective, let $a,a' \in A$ with $g \circ f(a) = g \circ f(a')$. Then $g(f(a)) = g(f(a'))$ and by injectivity of $g$ we find $f(a) = f(a')$. Injectivity of $f$ implies $a = a'$.
3. Proofs 1. and 2. imply 3. by definition of bijectivity.

Proposition: if f: A \to B and g: B \to A are maps with f \circ g = I_B and g \circ f = I_A, where I_A and I_B denote the identity maps on A and B, respectively. Then f and g are bijections and f^{-1} = g and g^{-1} = f.

??? note "Proof:"

Suppose $f A \to B$ and $g: B \to A$ are maps with $f \circ g = I_B$ and $g \circ f = I_A$. Let $b \in B$ then $f(g(b)) = b$, thus $f$ is surjective. If $a,a' \in A4 with $f(a) = f(a')$, then $a = g(f(a)) = g(f(a')) = a' and hence $f$ is injective. Therefore $f$ is bijective and by symmetry $g$ is also bijective.

Proposition: suppose f: A \to B and g: B \to C are bijective maps. Then the inverse of the map g \circ f equals f^{-1} \circ g^{-1}.

??? note "Proof:"

Suppose $f: A \to B$ and $g: B \to C$ are bijective maps. Then for all $a \in A$ we have $(f^{-1} \circ g^{-1}) (g \circ f)(a) = f^{-1}(g^{-1}(g(f(a)))) = f^{-1}(f(a)) = a$.