Composition of functions

Given two functions $f : A \to B$ and $g : B \to C$, the codomain of $f$ must match the domain of $g$ for the composite to be defined. The composite function $g \circ f : A \to C$ is defined by

where: $f$ is the inner function (applied first), $g$ is the outer function (applied second), $A$ is the domain of $f$, $B$ is the shared codomain-of-$f$/domain-of-$g$, and $C$ is the codomain of $g$. Read $g \circ f$ from right to left: "first $f$, then $g$." A few facts worth carrying forward without formal proof:

• Not commutative. $g \circ f$ and $f \circ g$ are generally different functions — sometimes one of the two isn't even defined.
• Associative. $(h \circ g) \circ f = h \circ (g \circ f)$ whenever every composite involved is defined, so a chain of three functions can be bracketed either way.
• Preserves one-one and onto. If $f$ and $g$ are both one-one, $g \circ f$ is one-one; if both are onto, $g \circ f$ is onto.

The picture below shows the chain of sets and how a single input $x$ threads through both machines.

Knowing the composite formula is half the story — the harder, exam-style subtlety is figuring out where it is actually valid. A wrong guess here is the single most common slip in this topic, so it deserves a careful derivation: we work out the general rule that pins down the domain of any composite function $f \circ g$, showing why simply intersecting the domains of $f$ and $g$ separately is not enough, and why you must instead trace which outputs of the inner function are allowed to land inside the outer function's domain.

Step 1. Start from the definition $(f \circ g)(x) = f(g(x))$. For the right-hand side to even be computable, two things must independently hold — $g(x)$ must exist, and then $f$ must accept the value $g(x)$ produces.

Step 2. The first requirement is simply that $x$ lies in the domain of $g$, since otherwise $g(x)$ is not defined and there is nothing to feed forward.

Step 3. The second requirement is that the output $g(x)$, not $x$ itself, must lie in the domain of $f$. This is the step people skip: they check whether $x$ is a valid input to $f$, but $f$ never sees $x$ directly — it only ever sees $g(x)$.

Step 4. Combining both conditions gives the domain of the composite as $\{x \in \text{Dom}(g) : g(x) \in \text{Dom}(f)\}$. This set can be smaller than the domain of $g$ alone (extra values get filtered out by the second condition) but it can never contain any $x$ outside the domain of $g$ (the first condition is a hard floor). Applying this to $f(x) = \sqrt{x}$ with domain $[0,\infty)$ and $g(x) = x - 4$ with domain $\mathbb{R}$: every real $x$ passes Step 2, but Step 3 demands $x - 4 \ge 0$, i.e. $x \ge 4$, so the composite domain shrinks to $[4, \infty)$.

Let $f(x) = \sqrt{x}$ with domain $[0,\infty)$, and $g(x) = x - 4$ with domain $\mathbb{R}$ — the same pair used above. Earlier we found $(f\circ g)(x) = \sqrt{x-4}$ with domain $[4,\infty)$. Now find $(g \circ f)(x)$ and its domain, and compare.

Step 1. Apply the definition with $f$ as the inner function: $(g \circ f)(x) = g(f(x)) = g(\sqrt{x})$.

Step 2. Substitute $\sqrt{x}$ into $g(t) = t - 4$: $(g \circ f)(x) = \sqrt{x} - 4$.

Step 3. Find the domain using the rule from the derivation. First, $x$ must lie in the domain of $f$, so $x \ge 0$. Second, the output $f(x) = \sqrt{x}$ must lie in the domain of $g$, which is all of $\mathbb{R}$ — so this second condition places no extra restriction.

Step 4. The domain of $g \circ f$ is therefore just $[0, \infty)$, unlike the domain of $f \circ g$, which was the smaller set $[4, \infty)$. So $(g \circ f)(x) = \sqrt{x} - 4$ with domain $[0, \infty)$, while $(f \circ g)(x) = \sqrt{x - 4}$ with domain $[4, \infty)$ — same two functions, opposite order, genuinely different formulas and different domains. This is the clearest possible illustration that composition is not commutative in either its rule or its domain.

• Writing $g \circ f$ but computing $f(g(x))$ instead of $g(f(x))$ — the function on the right of $\circ$ always acts first.
• Assuming $f \circ g = g \circ f$; as the worked example shows, both the formula and the domain can differ.
• Checking whether $x$ is a valid input to the outer function instead of checking whether the inner function's output is a valid input to it — this is the exact mistake Step 3 of the derivation guards against.
• Forgetting that the composite's domain can never exceed the inner function's domain, even if the algebra after simplification looks defined everywhere.

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