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12.2 Operation with functions

Composition

Given \(f: A \rightarrow B\) and \(g: B \rightarrow C\), we define.\((g \circ f)(a) = g(f(a)).\)

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Example

\(f(x) = x^2: \mathbb{R} \rightarrow \mathbb{R}\) \(g(x) = \sin(x): \mathbb{R} \rightarrow [-1, 1]\)

\(g \circ f: \mathbb{R} \rightarrow [-1, 1]\) is given by \(g \circ f(x) = \sin(x^2)\)

Some properties

  1. Given: \(f: A \to B\)

    • \(f \circ id_A = f\)
    • \(id_B \circ f = f\)

  2. Composition is associative: Given image​ then: \(h \circ (g \circ f) = (h \circ g) \circ f\)

    Proof: These are functions from \(A\) to \(D\).

    Given \(a\in A\) then \(h\circ(g\circ f)\left(a\right)=h(g\circ f(a))=h(g(f))\left(a\right)=\left(h\circ g\right)\circ f\left(a\right).\)​

Characteristic function

Restriction

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Given \(f: U \to B\) and a subset \(A\subset U\), the restriction to \(A\) is the function \(f|_A: A \to B\) given by \(f|_A(a) = f(a)\)

In other words, \(f|_A = f \circ i\), where \(i: A \to U\) is the inclusion.

Example:

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Extension

Given \(f: A \to B\) and \(U \supseteq A\), an extension of \(f\) to \(U\) is a function \(g: U \to B\) such that \(g|_A = f\)

In other words, \(g \circ i = f\).

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Example

\(f(x) = \sqrt{x}\), \(f:R_{\geq0}\to\mathbb{R}\)

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We want to extend it to \(\mathbb{R}\).

\(g: \mathbb{R} \to \mathbb{R}\), \(g(x)=\begin{cases}-x & \text{if }x<0\\ \sqrt{x} & \text{if }x\geq0\end{cases}\)

Example

\(f(x) = \frac{1}{x}\), \(f: \mathbb{R} - \{0\} \to \mathbb{R}\)

We define \(g: \mathbb{R} \to \mathbb{R}\) by \(g(x)=\begin{cases}\frac{1}{x} & \text{if }x\neq0\\ 0 & \text{if }x=0\end{cases}\)

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Remark

We can also extend the codomain or both.

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\(g\) is an extension of \(f\) if \(g(a) = f(a)\) for \(a \in A\).

Remark

We can also restrict the codomain.

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Injective functions

A function \(f: A \to B\) is said to be one-to-one or injective if:

\(\forall x_1, x_2 \in A, \, (f(x_1) = f(x_2) \implies x_1 = x_2)\).

Or

\(\forall x_1,x_2\in A,\,(x_1\neq x_2\Rightarrow f(x_1)\neq f(x_2))\).

We can also write: \(\forall (b \in B) \, f(x) = b\) has at most one solution in \(A\).

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Example

\(f(n)=n^2,f:\mathbb{N}\to\mathbb{N}\).

Given \(n_1, n_2 \in \mathbb{N}\), we can assume that \(n_1 < n_2\).

Then \(f(n_1)=n_1\cdot n_1<n_1\cdot n_2<n_2\cdot n_2=f(n_2)\implies f(n_1)\neq f(n_2)\).

Non-example

\(g(n) = n^2\) \(g: \mathbb{Z} \to \mathbb{Z}\)

\(g\) is not one-to-one because \(g(1) = 1^2 = g(-1)\)

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Surjective functions

We say that \(f: A \to B\) is surjective or onto \(B\) if:

\(\forall b\in B,f^{-1}(b)\neq\emptyset\)

or \(\text{Im} \, f = B\)

or \(\forall b\in B,\exists x\in A:f(x)=b\)

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Example

\(f(x)=x^2\,,f:\mathbb{R}\to\mathbb{R}_{\geq}0\).

This function is surjective because: given \(y\geq 0\), we choose \(x=\sqrt{y_{}}\) and then \(f(\sqrt{y})=(\sqrt{y})^2=y\).

Given \(y \geq 0\), \(f(x) = y\) has two solutions, \(\sqrt{y}\) and \(-\sqrt{y}\).

Non-example

\(g(x)=x^2,g:\mathbb{R}\to\mathbb{R}\) is not surjective because \(x^2 = -1\) doesn't have solutions.

Horizontal line test

\(f: D \subset \mathbb{R} \to \mathbb{R}\).

\(f\) is injective if each horizontal line meets the graph in at most one point.

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\(f\) is surjective if each horizontal line meets the graph in at least one point.

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General properties

  1. \(Id_{A}:A\to A\) is surjective and injective.

  2. If \(f: A \to B\) injective and \(g: B \to C\) injective then \(g \circ f: A \to C\) injective

  3. If \(f: A \to B\) surjective and \(g: B \to C\) surjective then \(g \circ f: A \to C\) surjective

Proof

  1. Take \(id(x_1)=id(x_2)\Rightarrow x_1=x_2\)​

    For any \(x\in A\), \(\exists x\in A\), \(id(x)=x\)

  2. Take \(g(f(x_1))=g(f(x_2))\), since \(g\) is injective, then \(f(x_1)=f(x_2)\)​

    Also since \(f\) is injective, then \(x_1=x_2\), thus \(g\circ f\) is injective

  3. We need to prove for any \(c\in C\), \(\exists a\in A:g(f(a))=c\)

    Since \(f\) is surjective, then for any \(b\in B,\exists a\in A:f(a)=b\)

    Since \(g\) is surjective, then for any \(c\in C,\exists b\in B:g(b)=c\)

    Then we have for any \(c\in C,\exists a\in A:g(f\left(a\right))=c\)

Bijective function

A function \(f: A \to B\) is said to be a one-to-one correspondence or bijective if it is both injective and surjective.

In other words:

\(\forall b\in B,\;\exists!a\in A:f(x)=b\)

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Example

\(f(x)=2x+1\;,f:\mathbb{R}\to\mathbb{R}\).

Given \(y \in \mathbb{R}\), we solve \(2x + 1 = y\).

\(x = \frac{y - 1}{2}\). Just one solution.

\(f\) is bijective.

Properties

  1. \(I_{A}:A\to A\) is bijective then:

  2. \(f:A\rightarrow B\) is bijective, \(g:B\rightarrow C\) is bijective then \(g\circ f:A\rightarrow C\) is bijective

Proposition

If \(f: A \to B\) is bijective, then \(f^{-1}\) is a function form \(B\) to \(A\) and \(f^{-1}\circ f=id_A\) and \(f\circ f^{-1}=id_B\)

Proof:

Let \(f^{-1}=\left\lbrace\left(b,a\right):afb\right\rbrace=\left\lbrace\left(b,a\right):f\left(a\right)=b\right\rbrace\)​

Because \(f\) is surjective, given \(b \in B\), let \(\exists a \in A\) such that \(f(a) = b\)

Then \(Dom(f^{-1})=B\)

Given \((b,a_1)\), \((b,a_2)\in f^{-1}\) then \(f(a_1)=b\wedge f(a_2)=b\).

So \(a_1 = a_2\) since \(f\) is injective.

Finally, given \(a \in A\), we have \(f(a)=b\), so \(f^{-1}(b)=a\). So \(f^{-1}f(a)=a\Rightarrow f^{-1}\circ f=id_A\)​

Similarly, given \(b \in B\), we choose \(a \in A\) such that \(f(a)=b,b=f\left(a\right)=f(f^{-1}(b))\Rightarrow f\circ f^{-1}=id_{B}\).

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