Composition of functions

Subsection 2.3.7 Composition of functions

Definition 2.3.7.1.

Let $f \colon A \to B\text{,}$ and let $g \colon B \to C\text{.}$ The composition, denoted $g \circ f\text{,}$ is the function $g \circ f \colon A \to C$ defined by

\begin{equation*} (g \circ f)(x) = g(f(x)). \end{equation*}

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$Three sets A and B and C are represented each by an oval. The ovals do not overlap. An arrow labeled f goes from A to B. An arrow labeled g goes from B to C. An arrow labeled the composition of g and f goes from A to C.$

Figure 2.3.7.2. Composition $g \circ f$ of two functions $f:A\to B$ and $g:B\to C\text{.}$

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$Three sets A and B and C are represented each by an oval. The ovals do not overlap. Inside A are the numbers 1, 2, and 3. Inside B are the numbers 2, 3, and 4. Inside C are the numbers 0, 1, and 2. An arrow labeled f goes from 1 in A to to 3 in B. An arrow labeled f goes from 2 in A to to 4 in B. An arrow labeled f goes from 3 in A to to 4 in B. An arrow labeled g goes from 2 in B to to 0 in C. An arrow labeled g goes from 3 in B to to 1 in C. An arrow labeled g goes from 4 in B to to 2 in C. An arrow labeled the composition of g and f goes from 1 in A to to 1 in C. An arrow labeled the composition of g and f goes from 2 in A to to 2 in C. An arrow labeled the composition of g and f goes from 2 in A to to 2 in C.$

Figure 2.3.7.3. Two functions $f:A\to B$ and $g:B\to C$ and their composition $g \circ f$ where $A=\{1,2,3\}\text{,}$ $B=\{2,3,4\}\text{,}$ and $C=\{0,1,2\}\text{.}$

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Example 2.3.7.4. Composition.

Let $f(x) = x^2 + 2x - 1$ and $g(x) = 3x - 5$ be functions from $\RR$ to $\RR\text{.}$

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Let’s compute some compositions. First, consider $f \circ g\text{.}$

\begin{align*} (f \circ g) (x) \amp = f(g(x)) \\ \amp = f(3x - 5) \\ \amp = (3x - 5)^2 + 2(3x - 5) - 1 \\ \amp = (9x^2 - 30x + 25) + (6x - 10) - 1 \\ \amp = 9x^2 - 24x + 14. \end{align*}

Next, let’s compose them in the opposite order.

\begin{align*} (g \circ f)(x) \amp = g(f(x)) \\ \amp =g(x^2 + 2x - 1) \\ \amp = 3(x^2 + 2x - 1) - 5 \\ \amp = (3x^2 + 6x - 3) - 5 \\ \amp = 3x^2 + 6x - 8. \end{align*}

Finally, let’s compose $g$ with itself.

\begin{align*} (g \circ g) (x) \amp = g(g(x)) \\ \amp = g(3x - 5) \\ \amp = 3(3x - 5) - 5 \\ \amp = (9x - 15) - 5 \\ \amp = 9x - 20. \end{align*}

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Definition 2.3.7.5.

Let $A$ be a nonempty set. We call $\id{A}:A\to A$ with $\id{A}(x)=x$ the identity function on $A\text{.}$

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It follows from Definition 2.3.6.7 and Definition 2.3.7.1 that for a bijective function $f:A\to B$ and its inverse $f^{-1}:B\to A$ that

\begin{equation*} f\circ f^{-1}= \id{B}\quad \text{ and }\quad f^{-1}\circ f=\id{A}. \end{equation*}

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