Composition of functions

If you can map $A$ to $B$ and then $B$ to $C$, you can map $A$ to $C$ by performing the two steps in sequence. That sequence is called the composition of the two functions, written $g \circ f$. Composition behaves like multiplication: it is associative, it has an identity, and on the set of bijections of any set $A$ to itself it even has inverses, producing what algebraists call the symmetric group $S_A$. For our purposes the immediate payoff is that composition is how we will define and verify inverse functions in the next subtopic.

Definition

Let $f : A \to B$ and $g : B \to C$ be functions. The composition $g \circ f : A \to C$ is defined by

$(g \circ f)(x) = g(f(x)) \quad \text{for all } x \in A.$

Read $g \circ f$ as "$g$ after $f$". Caution: $g \circ f$ requires the codomain of $f$ to be contained in the domain of $g$.

Three immediate properties

Associativity. If $f : A \to B$, $g : B \to C$, $h : C \to D$, then $h \circ (g \circ f) = (h \circ g) \circ f.$ Proof. Both sides take $x$ to $h(g(f(x)))$.

Identity. Let $I_A : A \to A$, $I_A(x) = x$. Then for any $f : A \to B$, $f \circ I_A = f \quad \text{and} \quad I_B \circ f = f.$

Non-commutativity. In general $f \circ g \neq g \circ f$, even when both make sense. For $f(x) = x + 1$ and $g(x) = 2x$, $f(g(x)) = 2x + 1$ but $g(f(x)) = 2x + 2$.

Preservation theorems

Theorem. Let $f : A \to B$ and $g : B \to C$.

1. If $f$ and $g$ are both one-one, then $g \circ f$ is one-one.
2. If $f$ and $g$ are both onto, then $g \circ f$ is onto.
3. If $f$ and $g$ are both bijective, then $g \circ f$ is bijective.

Proof of (1). Suppose $(g \circ f)(x_1) = (g \circ f)(x_2)$. Then $g(f(x_1)) = g(f(x_2))$. Since $g$ is one-one, $f(x_1) = f(x_2)$. Since $f$ is one-one, $x_1 = x_2$.

Proof of (2). Let $z \in C$. Since $g$ is onto, there is $y \in B$ with $g(y) = z$. Since $f$ is onto, there is $x \in A$ with $f(x) = y$. Then $(g \circ f)(x) = z$. $\blacksquare$

A subtler theorem

Theorem. Let $f : A \to B$ and $g : B \to C$.

1. If $g \circ f$ is one-one, then $f$ is one-one.
2. If $g \circ f$ is onto, then $g$ is onto.

Proof of (1). Suppose $f(x_1) = f(x_2)$. Apply $g$: $g(f(x_1)) = g(f(x_2))$, i.e., $(g \circ f)(x_1) = (g \circ f)(x_2)$. Since $g \circ f$ is one-one, $x_1 = x_2$.

Proof of (2). Let $z \in C$. Since $g \circ f$ is onto, there is $x$ with $g(f(x)) = z$. Setting $y = f(x) \in B$, $g(y) = z$. $\blacksquare$

Note the asymmetry: the hypothesis tells us about the outer map in one case and the inner in the other.

A common JEE trap

The converse of the above is false. $g \circ f$ being one-one does not force $g$ to be one-one , only $f$. Symmetrically, $g \circ f$ onto does not force $f$ onto. The standard counter-example uses $A = \{1\}$, $B = \{a, b\}$, $C = \{x\}$ with $f(1) = a$ and $g(a) = g(b) = x$.

Worked examples

Example 1. $f(x) = 3x + 1$, $g(x) = x^2$. Find $f \circ g$ and $g \circ f$.

$(f \circ g)(x) = f(x^2) = 3x^2 + 1$. $(g \circ f)(x) = g(3x + 1) = (3x + 1)^2 = 9x^2 + 6x + 1$.

Example 2. $f : \mathbb{R} \to \mathbb{R}$, $f(x) = |x|$ and $g(x) = 5x - 2$. Compute $(f \circ g)(3)$ and $(g \circ f)(-3)$.

$(f \circ g)(3) = f(13) = 13$. $(g \circ f)(-3) = g(3) = 13$. (Equal here by coincidence.)

Example 3. Let $f(x) = \tfrac{1}{1 - x}$ for $x \neq 1$. Compute $f \circ f$ and $f \circ f \circ f$.

$(f \circ f)(x) = f(\tfrac{1}{1 - x}) = \tfrac{1}{1 - \tfrac{1}{1 - x}} = \tfrac{1 - x}{(1 - x) - 1} = \tfrac{1 - x}{-x} = \tfrac{x - 1}{x}$.

$(f \circ f \circ f)(x) = f(\tfrac{x - 1}{x}) = \tfrac{1}{1 - \tfrac{x - 1}{x}} = \tfrac{x}{x - (x - 1)} = x$.

So $f^3 = I$ on its domain. A function of period $3$ under composition.

Example 4. $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x + 7$. $g : \mathbb{R} \to \mathbb{R}$, $g(x) = x - 7$. Compute $f \circ g$ and $g \circ f$.

Both equal $x$, i.e., the identity. This is the prototype of an inverse pair.

Example 5. Show $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ for bijections $f$ and $g$.

If $f$ and $g$ are bijections, so is $g \circ f$, by the preservation theorem. Compute:

$(f^{-1} \circ g^{-1}) \circ (g \circ f) = f^{-1} \circ (g^{-1} \circ g) \circ f = f^{-1} \circ I \circ f = f^{-1} \circ f = I.$

Similarly on the other side. So $f^{-1} \circ g^{-1}$ is the inverse of $g \circ f$.

The reversal of order is the most-tested formula in this chapter.

Example 6. Let $f, g : \mathbb{R} \to \mathbb{R}$ with $f(x) = \sin x$ and $g(x) = x^2$. Compute the four possible compositions.

$(f \circ g)(x) = \sin(x^2)$, $(g \circ f)(x) = (\sin x)^2 = \sin^2 x$, $(f \circ f)(x) = \sin(\sin x)$, $(g \circ g)(x) = x^4$.

Try it yourself

1. $f(x) = 2x - 3, g(x) = (x + 3)/2$. Compute $f \circ g$ and $g \circ f$.
2. $f(x) = x^2 + 1, g(x) = \sqrt{x - 1}$. State the domains where $f \circ g$ and $g \circ f$ are defined.
3. Find a function $f$ with $f \circ f = I$ on $\mathbb{R}$. (Hint: try $f(x) = -x$ or $f(x) = c - x$.)
4. Prove $f \circ I_A = f$ for any $f : A \to B$.
5. If $f(x) = ax + b$, find conditions on $a, b$ so that $f \circ f = I$.
6. Let $f(x) = \tfrac{x}{x - 1}$ on $\mathbb{R} \setminus \{1\}$. Compute $f \circ f$.
7. Show: if $f$ and $g$ both surjective from $A$ to $A$, then $g \circ f$ is surjective.
8. Provide a counter-example: $g \circ f$ surjective but $f$ not surjective.
9. Compute $(g \circ f)(3)$ and $(f \circ g)(3)$ where $f(x) = x + 2, g(x) = x^2$.
10. Given $f : \mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = x^2 + x + 1$, find $f(0)$.
11. Prove associativity of composition rigorously by expanding each side at an arbitrary input.
12. Find $f$ on $\mathbb{R}$ such that $f \circ f \circ f = I$, but $f \neq I$.
13. Let $f, g : A \to A$ with $f \circ g = I_A$. Show $f$ is onto and $g$ is one-one.
14. If $f \circ g = g \circ f$ for all $g$, what can $f$ be?

Pitfalls / Tricks

• Reading order matters: $g \circ f$ means apply $f$ first, then $g$. The notation reads right-to-left.
• Composition is associative but never assume commutativity unless proved.
• For finite sets the iteration $f^n = f \circ f \circ \dots \circ f$ ($n$ times) is eventually periodic; in JEE it is common to ask for $f^{100}$ in disguise.
• The reversal formula $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ is the same pattern as $(AB)^{-1} = B^{-1} A^{-1}$ for matrices, and is also a daily formula in the matrices chapter.

Once you trust composition, inverses become a one-line definition.