Composition of Functions

Chaining functions

When the output set of one function feeds the input set of another, the two can be chained.

Definition. Let $f : A \to B$ and $g : B \to C$. The composition of $f$ and $g$, written $g \circ f$ (read "$g$ of $f$"), is the function $g \circ f : A \to C$ given by $$(g \circ f)(x) = g\big(f(x)\big), \quad \forall\, x \in A$$

Read it right to left: $f$ acts first, then $g$ acts on the result. The middle set $B$ must match — $f$'s outputs must be legal inputs for $g$.

Worked chain on finite sets. Let $f : \{2,3,4,5\} \to \{3,4,5,9\}$ with $f(2) = 3$, $f(3) = 4$, $f(4) = f(5) = 5$, and $g : \{3,4,5,9\} \to \{7,11,15\}$ with $g(3) = g(4) = 7$, $g(5) = g(9) = 11$. Then $$(g \circ f)(2) = g(3) = 7, \quad (g \circ f)(3) = g(4) = 7, \quad (g \circ f)(4) = g(5) = 11, \quad (g \circ f)(5) = g(5) = 11$$

Order matters: $g \circ f \neq f \circ g$ in general

For $f(x) = \cos x$ and $g(x) = 3x^2$ on $\mathbb{R}$: $$(g \circ f)(x) = g(\cos x) = 3\cos^2 x, \qquad (f \circ g)(x) = f(3x^2) = \cos(3x^2)$$ At $x = 0$: $\ (g \circ f)(0) = 3\cos^2 0 = 3$ while $(f \circ g)(0) = \cos 0 = 1$. One disagreement is enough: $g \circ f \neq f \circ g$. Composition is not commutative — but it is associative: $h \circ (g \circ f) = (h \circ g) \circ f$ whenever the chains make sense, because both sides send $x$ to $h(g(f(x)))$.

How composition interacts with one-one and onto

Two facts that exams (especially JEE) probe:

Fact 1. If $g \circ f$ is one-one, then $f$ must be one-one (a collision in $f$ would survive into $g \circ f$) — but $g$ need not be.

Fact 2. If $g \circ f$ is onto, then $g$ must be onto (everything in $C$ is reached through $g$) — but $f$ need not be.

And in the convenient direction: the composite of two one-one functions is one-one; the composite of two onto functions is onto; hence the composite of two bijections is a bijection.

Invertible Functions

Undoing a function

Definition. A function $f : X \to Y$ is invertible if there exists a function $g : Y \to X$ such that $$g \circ f = I_X \quad \text{and} \quad f \circ g = I_Y$$ where $I_X$ and $I_Y$ are the identity functions. The function $g$ is called the inverse of $f$, written $f^{-1}$.

Both equations matter: $g$ must undo $f$ from the left and from the right.

The central theorem

Theorem. $f$ is invertible if and only if $f$ is one-one and onto (bijective).

Why: if $f$ has an inverse, then $f$ cannot collide (the inverse could not decide where to send the shared image) and cannot miss any $y$ (the inverse must send every $y$ somewhere that maps back to it). Conversely a bijection pairs $X$ and $Y$ perfectly, and "follow the pairing backwards" defines the inverse.

This theorem is a labour-saving device: to prove invertibility you may simply prove one-one and onto — no formula for $f^{-1}$ required. When the formula is required, use the template below.

Computing an inverse (the template)

Step 1 — set $y = f(x)$ and solve for $x$ in terms of $y$.

Step 2 — the solution defines $g(y)$; state its domain (the co-domain of $f$).

Step 3 — verify both compositions: $g(f(x)) = x$ and $f(g(y)) = y$.

Worked inverse. $f : \mathbb{N} \to Y$, $f(x) = 4x + 3$, where $Y = \{y \in \mathbb{N} : y = 4x + 3 \text{ for some } x \in \mathbb{N}\}$.

Step 1: $y = 4x + 3 \Rightarrow x = \dfrac{y - 3}{4}$.

Step 2: define $g : Y \to \mathbb{N}$ by $g(y) = \dfrac{y - 3}{4}$ (a natural number by the definition of $Y$).

Step 3: $g(f(x)) = \dfrac{(4x + 3) - 3}{4} = x$ and $f(g(y)) = 4 \cdot \dfrac{y - 3}{4} + 3 = y$. Both identities hold, so $f$ is invertible with $f^{-1}(y) = \dfrac{y - 3}{4}$. $\blacksquare$

Note how the co-domain was chosen to be exactly the range — that choice is what made $f$ onto and hence invertible.

Properties worth memorising

$$\big(f^{-1}\big)^{-1} = f, \qquad (g \circ f)^{-1} = f^{-1} \circ g^{-1} \ \ (\text{socks-shoes rule: undo in reverse order})$$

A function can even be its own inverse (a self-inverse function): $f(x) = \dfrac{4x + 3}{6x - 4}$, $x \neq \frac{2}{3}$, satisfies $(f \circ f)(x) = x$, so $f^{-1} = f$. Other self-inverse examples: $f(x) = -x$, $\ f(x) = \frac{1}{x}$, $\ f(x) = a - x$.

Common mistakes to avoid

Mistake 1 — $f^{-1}(x)$ vs $\frac{1}{f(x)}$. The inverse function is not the reciprocal. $f(x) = 2x$ has $f^{-1}(x) = \frac{x}{2}$, not $\frac{1}{2x}$.

Mistake 2 — inverting a non-bijection. $f(x) = x^2$ on $\mathbb{R}$ has no inverse; only after restricting to $[0, \infty)$ (making it bijective onto $[0, \infty)$) does $\sqrt{x}$ become its inverse.

Mistake 3 — checking only one composition. $g \circ f = I_X$ alone does not prove invertibility in general; the definition demands both $g \circ f = I_X$ and $f \circ g = I_Y$.

Mistake 4 — wrong order in $(g \circ f)^{-1}$. It is $f^{-1} \circ g^{-1}$, not $g^{-1} \circ f^{-1}$ — you take off shoes before socks.

Solved Examples

Example 1 — Composing along finite tables

Let $f : \{2, 3, 4, 5\} \to \{3, 4, 5, 9\}$ and $g : \{3, 4, 5, 9\} \to \{7, 11, 15\}$ be defined by $f(2) = 3$, $f(3) = 4$, $f(4) = f(5) = 5$, and $g(3) = g(4) = 7$, $g(5) = g(9) = 11$. Find $g \circ f$.

Step 1 — push each element through both tables: $$(g \circ f)(2) = g(f(2)) = g(3) = 7$$ $$(g \circ f)(3) = g(f(3)) = g(4) = 7$$ $$(g \circ f)(4) = g(f(4)) = g(5) = 11$$ $$(g \circ f)(5) = g(f(5)) = g(5) = 11$$

Answer: $g \circ f$ sends $2 \mapsto 7$, $3 \mapsto 7$, $4 \mapsto 11$, $5 \mapsto 11$. (Note $15$ in the final co-domain is never reached — composition can lose surjectivity.)

Example 2 — Order matters

Find $g \circ f$ and $f \circ g$ for $f(x) = \cos x$ and $g(x) = 3x^2$ on $\mathbb{R}$, and show $g \circ f \neq f \circ g$.

Step 1 — compose both ways: $$(g \circ f)(x) = g(\cos x) = 3\cos^2 x, \qquad (f \circ g)(x) = f(3x^2) = \cos(3x^2)$$

Step 2 — find one point of disagreement: at $x = 0$, $\ 3\cos^2 0 = 3$ but $\cos 0 = 1$.

Answer: $g \circ f \neq f \circ g$ — composition is not commutative, and a single differing value proves it.

Example 3 — A composition that simplifies

If $f : \mathbb{R} \to \mathbb{R}$ is $f(x) = 8x^3$ and $g : \mathbb{R} \to \mathbb{R}$ is $g(x) = x^{1/3}$, find $g \circ f$ and $f \circ g$.

Step 1 — $g \circ f$: $$(g \circ f)(x) = g(8x^3) = (8x^3)^{1/3} = 2x$$

Step 2 — $f \circ g$: $$(f \circ g)(x) = f(x^{1/3}) = 8\big(x^{1/3}\big)^3 = 8x$$

Answer: $(g \circ f)(x) = 2x$ and $(f \circ g)(x) = 8x$ — both linear, but different: yet another reminder that the order of composition matters.

Example 4 — Proving invertibility and finding the inverse

Let $f : \mathbb{N} \to Y$, $f(x) = 4x + 3$, where $Y = \{y \in \mathbb{N} : y = 4x + 3$ for some $x \in \mathbb{N}\}$. Show that $f$ is invertible and find its inverse.

Step 1 — solve $y = f(x)$ for $x$: $y = 4x + 3$ gives $x = \dfrac{y - 3}{4}$, a natural number for every $y \in Y$ by the very definition of $Y$.

Step 2 — define the candidate inverse: $g : Y \to \mathbb{N}$, $\ g(y) = \dfrac{y - 3}{4}$.

Step 3 — verify both compositions: $$g(f(x)) = \frac{(4x + 3) - 3}{4} = x, \qquad f(g(y)) = 4\left(\frac{y - 3}{4}\right) + 3 = y$$

Answer: $g \circ f = I_{\mathbb{N}}$ and $f \circ g = I_Y$, so $f$ is invertible with $f^{-1}(y) = \dfrac{y - 3}{4}$.

Example 5 — Another linear inverse

Let $f : \mathbb{R} \to \mathbb{R}$ be $f(x) = 10x + 7$. Find the function $g : \mathbb{R} \to \mathbb{R}$ such that $g \circ f = f \circ g = I_{\mathbb{R}}$.

Step 1 — solve: $y = 10x + 7 \Rightarrow x = \dfrac{y - 7}{10}$, so define $g(y) = \dfrac{y - 7}{10}$.

Step 2 — verify: $$g(f(x)) = \frac{(10x + 7) - 7}{10} = x, \qquad f(g(y)) = 10 \cdot \frac{y - 7}{10} + 7 = y$$

Answer: $g(y) = \dfrac{y - 7}{10}$ — i.e. $f^{-1}$ reverses "multiply by 10, add 7" as "subtract 7, divide by 10", undoing the steps in reverse order.

Example 6 — A self-inverse function

Show that $f(x) = \dfrac{4x + 3}{6x - 4}$, defined for $x \neq \frac{2}{3}$, satisfies $(f \circ f)(x) = x$. What is $f^{-1}$?

Step 1 — compute $f(f(x))$ over a common denominator: $$f(f(x)) = \frac{4 \cdot \frac{4x+3}{6x-4} + 3}{6 \cdot \frac{4x+3}{6x-4} - 4} = \frac{\frac{16x + 12 + 18x - 12}{6x-4}}{\frac{24x + 18 - 24x + 16}{6x-4}}$$

Step 2 — simplify: the $(6x - 4)$'s cancel, leaving $$f(f(x)) = \frac{34x}{34} = x$$

Answer: $f \circ f$ is the identity, so $f$ is invertible and $f^{-1} = f$ — a self-inverse function. Applying it twice always returns the starting value.

Example 7 — Using the bijection theorem instead of a formula

Show that the parity-swap function $f : \mathbb{N} \to \mathbb{N}$, $f(x) = x + 1$ for odd $x$ and $f(x) = x - 1$ for even $x$, is invertible, and find $f^{-1}$.

Step 1 — $f$ is bijective (proved in the previous section: it pairs $1 \leftrightarrow 2$, $3 \leftrightarrow 4, \ldots$), so by the theorem it is invertible.

Step 2 — find the inverse by undoing the pairing: the inverse must send $2 \mapsto 1, 1 \mapsto 2, 4 \mapsto 3, 3 \mapsto 4, \ldots$ — which is exactly what $f$ itself does.

Answer: $f^{-1} = f$: swapping each odd number with its even successor is its own undo. (Verify: $f(f(x)) = x$ — an odd $x$ goes to $x + 1$, even, which returns to $x$.)