Proof of the associativity of composition

Proof

We compare the two functions at each point.

1. Let $f \colon A \to B$, $g \colon B \to C$ and $h \colon C \to D$ be functions.
2. By the definition of composition, $g \circ f \colon A \to C$ and $h \circ g \colon B \to D$. So $h \circ (g \circ f)$ and $(h \circ g) \circ f$ are both functions from $A$ to $D$.
3. Let $a \in A$. By the definition of composition, applied twice, $(h \circ (g \circ f))(a) = h((g \circ f)(a)) = h(g(f(a)))$.
4. Also $((h \circ g) \circ f)(a) = (h \circ g)(f(a)) = h(g(f(a)))$.
5. By steps 3 and 4, the two functions take the same value at each $a \in A$.
6. A function $u \colon A \to D$ is the set of pairs $(a, u(a))$ with $a \in A$. By step 5, both functions are the same set of pairs.
7. Two sets with the same elements are equal, by the axiom of extensionality. Hence $h \circ (g \circ f) = (h \circ g) \circ f$.