0.3: Functions

 

Definition:  Cartesian Product

Let \(A\) be a set. Then \(A \times A= \{(a_1,a_2)| a_1, a_2 \in A \) \} is the Cartesian product of \(A\).

Example \(\PageIndex{1}\)

If \(A = \mathbb{R}\) then this is the Cartesian plane -  Like \(\{x,y: x,y \in A\}\).

Definition: Function

A function \(f\) is a subset (symbol is \(\subseteq\)) of \(A \times B\) such that for every \(a \in A\) there exists a unique \(b \in B\) such that \(f(a)=b\). 

Denoted by \(f: A \rightarrow B\) or \(A \xrightarrow{f} B\).

\(A\) is the domain of the function \(f\).  

\(B\) is the co-domain of the function \(f\).

The range of \(f\) is \(f(A)=\big{\{}f(x): x \in A \big{\}} \subseteq B\)

A well-defined function is one that for every value of \(A\), there is one and only one value of \(B\).

 

Definition: 

Let \(f\) be a function from \(A \rightarrow B\).

1. \(f\) is called onto (surjective) if \(f(A)=B\).

2. \(f\) is called one to one (injective) if \(f(x_1)=f(x_2), x_1,x_2 \in A \) then \(x_1=x_2\).

3. \(f\) is bijective if it is surjective and injective.

4. A function can be injective, surjective, bijective (both injective & surjective), or neither.

 Note: 

Let \(f: A \rightarrow B\).  If \(f\) is 1-1 and onto then \(f\) is a bijection from \(A \rightarrow B.\) 

If there is a bijection between two finite sets, the number of elements in each set is the same.  Notationally:  |A| = |B|.

Example \(\PageIndex{2}\)

Example \(\PageIndex{3}\)

Consider the mapping:

Is this mapping 

a) Well defined?

b)  Is it surjective? 

c) Is it injective? 

d) Is it bijective?

Solution

The function is well-defined. Not injective. Not subjective, not bijective.

 

Example \(\PageIndex{4}\)

Consider the mapping:

Is this mapping 

a) Well defined?

b)  Is it surjective? 

c) Is it injective? 

d) Is it bijective?

Solution

This is well defined bijective function.

Example \(\PageIndex{5}\)

Given \(f: \mathbb{Q} \rightarrow \mathbb{Z}\) where \(\mathbb{Q}=\{\frac{a}{b}:a,b \in \mathbb{Z}\}\) and \(f(\frac{a}{b})=a\).  Is this a well-defined function?  

Solution

Let \(A\) and \(B\) be sets such that \(A=\{\frac{a}{b}:a,b \in \mathbb{Z}\}\) and \(B=\mathbb{Z}\).

Using a counterexample, \(f(\frac{a}{b})\) is not a well-defined function since \(f(\frac{1}{2})=1\) and \(f(\frac{2}{4})=2\), but \(\frac{1}{2}=\frac{2}{4}\).  This is an example of an item in \(A\) that maps to more than one element in \(B\).

 

Example \(\PageIndex{6}\)

Let

 \(\mathbb{Q}=\big{\{}\frac{a}{b}: b\ne 0,\text{and } gcd(a,b)=1 \big{\}}\)

and \(f: \mathbb{Q} \rightarrow \mathbb{Z}\) that is defined by

 \(f(\frac{a}{b})=a.\)

 a) Is \(f\) a well-defined function? b)  Is it surjective?  c) Is it injective?  d) Is it bijective?

Solution

1. The function is well-defined.

Let \(A\) and \(B\) be set.

Let \(a \in A\).

We need to show that if \(f(a)=b\) then \(b\in B\) and if \(f(a)=b_1\) that \(b=b_1\).

We are given \(f: \mathbb{Q} \rightarrow \mathbb{Z}\), which means that this is a function thus \(f(a)\) will map to one and only one element of \(B\).  Thus if \(f(a)=b\) and if \(f(a)=b_1\) then \(b=b_1\).

In b) below, we have shown that the function is surjective, thus \(f(a)=b\),

 

2. The function is surjective

   Proof:

We shall show that \(f(\mathbb{Q})=\mathbb{Z}\) by showing that  \(f(\mathbb{Q}) \subseteq \mathbb{Z}\) and  \(\mathbb{Z} \subseteq f(\mathbb{Q})\).

Since we are given \(f: \mathbb{Q} \rightarrow \mathbb{Z}\),  \(f(\mathbb{Q}) \subseteq \mathbb{Z}\).

Next, we shall show that \(\mathbb{Z} \subseteq f(\mathbb{Q})\).

Let \(n \in \mathbb{Z}\), then \(\frac{n}{1} \in \mathbb{Q}\) and \(f\big(\frac{n}{1}\big)=n\).

Hence \(n \in f(\mathbb{Q})\).

Hence \(\mathbb{Z} \subseteq f(\mathbb{Q})\).

Since \(\mathbb{Z} \subseteq f(\mathbb{Q})\) and \(f(\mathbb{Q}) \subseteq \mathbb{Z}, f(\mathbb{Q})=\mathbb{Z}\).◻

 

3. The function is not injective.

   Counterexample:

Let \(A=\big{\{}\frac{a}{b}: b\ne 0,\text{and } gcd(a,b)=1 \big{\}}\) and \(B=\mathbb{Z}\).

Note \(\frac{1}{4},  \frac{1}{3} \in A\) and that \(f(\frac{1}{4}),  f(\frac{1}{3}) \in B\) and that \(f(\frac{1}{4})=f(\frac{1}{3})=1\). 

Since \(\frac{1}{4} \ne \frac{1}{3}\) , the function is not injective.

 

4. The function is not bijective since it has been shown not to be both injective and surjective.

 

Operation on functions: Composition

 

Definition:Composition of functions

Let \(f: A \rightarrow B\) and \(g: B \rightarrow C\). Then the composition functions of \(g\) and \(f\), \(g \circ f: A \to C \), defined by 

\((g \circ f)(a)=g(f(a)), \; \forall a \in A\).

Definition: Identity function

Identity function \(id_A: A \rightarrow A\) defined by \(id_A(x)=x, \forall x \in A\).

Definition: Inverse of a function

Let \(f: A \rightarrow B\) then \(f\) has an inverse \(g: B \rightarrow A\) if \((f \circ g)= id_B\) and \((g \circ f)=id_A\).  Note the domain of \(f \circ g\) is the domain of \(g\).

Let \(f:A \rightarrow B\) be a bijection.  Then the inverse of \(f\), denoted by \(f^{-1}:B \rightarrow A\) defined by \(f^{-1}(b)=a\) if \(f(a)=b\) if \(a \in A\) and \(b \in B\). In this case, \(f \circ f^{-1}=id_A, \) and \(f^{-1}\circ f =id_B.\)

 

 

Example \(\PageIndex{7}\)

Let \(f: A \rightarrow B\) and \(g: B \rightarrow C\). Then \(g \circ f: A \to C.\) 

1. If \(f\) and \(g\) are surjective, then \(g \circ f\) is surjective.

Proof:

We will show that \(g \circ f\) is surjective.

Let \(c \in C\) . Since g is surjective, there exists \(b \in B\)  such that \(g(b)=c.\)

Since f is surjective, there exists \(a \in A\)  such that \(f(b)=a\).

Hence there exists \(a \in A\) such that \(gf(f(a))=c.\)

Therefore, if \(f\) and \(g\) are surjective, then \(g \circ f\)  is surjective.◻

(better proof) If \(f\) is onto and \(g\) is onto then \(g \circ f\) is onto.

Proof:

We shall show that \((g \circ f)(A)=C\).

Since \(f\) and \(g\) are onto, by definition of a function, \((g \circ f)(A) \subseteq C\).

Now we need to show that \(C \subseteq (g \circ f)(A)\).

Let \(c \in C\).

Since \(g\) is onto \(\exists b \in B\) s.t. \(c=g(b)\).

Since \(f\) is onto, \(\exists a \in A\) s.t. \(b=f(a)\).

Consider \((g \circ f)(a)=g(f(a))\) (by definition)

        \(=g(b)\)

        \(=c\).

Therefore \(c \in (g \circ f)(A)\).

Hence \(C \subseteq (g \circ f)(A)\).

Since \((g \circ f)(A) \subseteq C\) and \(C \subseteq (g \circ f)(A)\), then \(g \circ f\) is onto.◻

 

Example \(\PageIndex{8}\)

1. If \(f\) and \(g\) are injective, then \(g \circ f\) is injective. That is,

If \(f\) is 1-1 and \(g\) is 1-1 then \(g \circ f\) is 1-1.

Proof:

Let   \(a_1,a_2 \in A\)  such that \(g \circ f (a_1)= g\circ f(a_2).\)  Then \(g(f(a_1))=g(f(a_2)).\)

Since  \(g\) is 1-1, \(f(a_1)=f(a_2)\).

Since,  \(f\) is 1-1, \(a_1=a_2.\)

 

Hence \(g \circ f\)  is injective.

 

 

Example \(\PageIndex{9}\)

Given \(A\) is a matrix and \(T_A:\mathbb{R}^2 \rightarrow \mathbb{R}^2\) defined by \(T_A \begin{bmatrix} x\\
y\end{bmatrix} = \begin{bmatrix} 1 &1\\
0 & 1\end{bmatrix} \begin{bmatrix} x\\
y\end{bmatrix}\).  Note this is a matrix transformation which is linear.

1. Is \(T_A\) 1-1?

Yes, \(T_A\) is 1-1.

Proof:

Let \(\begin{bmatrix} x_1\\
y_1\end{bmatrix}\)and \(\begin{bmatrix} x_2\\
y_2\end{bmatrix} \in \mathbb{R}^2\) s.t. \(T_A \begin{bmatrix} x_1\\
y_1\end{bmatrix} =T_A \begin{bmatrix} x_2\\
y_2\end{bmatrix}\).

We will show that \(\begin{bmatrix} x_1\\
y_1\end{bmatrix} =\begin{bmatrix} x_2\\
y_2\end{bmatrix}\).

\(T_A \begin{bmatrix} x\\
y\end{bmatrix} =\begin{bmatrix} x+y\\
y\end{bmatrix}\).

Thus  \(\begin{bmatrix} x_1 +y_1\\
y_1\end{bmatrix} =\begin{bmatrix} x_2+y_2\\
y_2\end{bmatrix}\).

Thus \(x_1+y_1=x_2+y_2\) and \(y_1=y_2\).

Thus \(x_1=x_2\) and \(y_1=y_2\).

Hence \(\begin{bmatrix} x_1\\
y_1\end{bmatrix} =\begin{bmatrix} x_2\\
y_2\end{bmatrix}\).

Thus \(T_A\) is 1-1.◻

 

2. Is \(T_A \begin{bmatrix} x\\
   y\end{bmatrix} \) onto?

Yes, \(T_A \begin{bmatrix} x\\
y\end{bmatrix} \) is onto.

Proof:

We know by definition that \(T_A \begin{bmatrix} x\\
y\end{bmatrix} \subseteq \mathbb{R}^2\).

We need to show that \(\mathbb{R}^2 \subseteq T_A \begin{bmatrix} x\\
y\end{bmatrix}\).

Let \( \begin{bmatrix} a\\
b\end{bmatrix} \in \mathbb{R}^2\).

Find \( \begin{bmatrix} x\\
y\end{bmatrix} \in \mathbb{R}^2\) s.t. \(\begin{bmatrix} 1 &1\\
0 & 1\end{bmatrix} \begin{bmatrix} x\\
y\end{bmatrix}=\begin{bmatrix} a\\
b\end{bmatrix}\).

\(\begin{bmatrix} x+y\\
y\end{bmatrix} =\begin{bmatrix} a\\
b\end{bmatrix}\).

Thus \(y=b\) and \(x=a-b\).

Thus \(T_A \begin{bmatrix} a-b\\
b\end{bmatrix} =\begin{bmatrix} a\\
b\end{bmatrix}\).

Thus \(T_A\) is onto.◻

 

3. Does \(T_A\) have an inverse?  If so, what is it?

a) Yes \(T_A\) has an inverse since it is a bijection (1-1 and onto) ∴ \(T_A^{-1}\) exists.

Proof:  By definition?  Obvious?

b) \(T_A^{-1}: \mathbb{R}^2 \rightarrow \mathbb{R}^2\).

\(T_A^{-1} \begin{bmatrix} x\\
y\end{bmatrix} =\begin{bmatrix} 1 & -1\\
0 & 1\end{bmatrix} \begin{bmatrix} x\\
y\end{bmatrix}\)

    \(=\begin{bmatrix} x-y\\
y\end{bmatrix} \).

Check: \( T_A^{-1} \cdot T_A=\begin{bmatrix} 1 & -1\\
0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1\\
0 & 1\end{bmatrix} =\begin{bmatrix} 1 & 0\\
0 & 1 \end{bmatrix}\).

Example \(\PageIndex{10}\)

Let \(f: A \rightarrow B\) and \(g: B \rightarrow C\). Then \(g \circ f:A \to C\)  . Given \(g \circ f\) is injective and \(f\) is onto, prove that \( g\) is injective.

Proof:

Assume \(g \circ f\) is injective.  (by definition)

Let \(x_1, x_2 \in B\) s.t. \(g(x_1)=g(x_2)\).

We shall show that \(x_1=x_2\).

Let \(a_1, a_2 \in A\).

Since \(f\) is surjective (by definition), we know that \(f(a_1)=x_1\) and \(f(a_2)=x_2\).

Since \(x_1 \in B\) and \(f\) is surjective, \(\exists a_1 \in A\) s.t. \(f(a_1)=x_1\).

Since \(x_2 \in B\) and \(f\) is surjective, \(\exists a_2 \in A\) s.t. \(f(a_2)=x_2\).

Since \(g(x_1)=g(x_2)\), then \(g(f(a_1))=g(f(a_2))\) and \(g \circ f)(a_1)=(g \circ f)(a_2)\).

Since \(g \circ f)\) is injective, \(a_1=a_2\).

Thus \(f(a_1)=f(a_2)\) since \(f\) is a function.

Therefore, \(x_1=x_2\).

Hence \(g\) is injective.