Answer
Yes, function composition is always associative, so \(\;h \circ (g \circ f) = (h \circ g)\circ f\).
Work Step by Step
## Explanation/Proof
Given three functions
\[
f: W \to X,\quad
g: X \to Y,\quad
h: Y \to Z,
\]
we consider the two compositions:
1. \(\bigl(h \circ (g \circ f)\bigr)(w)\).
2. \(\bigl((h \circ g)\circ f\bigr)(w)\).
Pick any \(w \in W\). Then:
- **Compute** \(\bigl(h \circ (g \circ f)\bigr)(w)\):
\[
(g \circ f)(w) \;=\; g\bigl(f(w)\bigr) \;\in\; Y,
\quad\text{so}\quad
\bigl(h \circ (g \circ f)\bigr)(w)
\;=\;
h\bigl((g \circ f)(w)\bigr)
\;=\;
h\bigl(g(f(w))\bigr).
\]
- **Compute** \(\bigl((h \circ g)\circ f\bigr)(w)\):
\[
f(w) \;\in\; X,
\quad\text{so}\quad
(h \circ g)\bigl(f(w)\bigr)
\;=\;
h\bigl(g(f(w))\bigr).
\]
Hence
\[
\bigl((h \circ g)\circ f\bigr)(w)
\;=\;
h\bigl(g(f(w))\bigr).
\]
Because both expressions equal \(h\bigl(g(f(w))\bigr)\) for every \(w\in W\), the two composite functions are identical:
\[
h \circ (g \circ f) = (h \circ g)\circ f.
\]
This property—called **associativity**—always holds for function composition, so there is no counterexample.
