Answer
True.** For all subsets \(C \subseteq Z\),
\[
(g \circ f)^{-1}(C)
\;=\;
f^{-1}\bigl(g^{-1}(C)\bigr).
\] This is a standard property of inverse images under compositions. Below is a short justification.
Work Step by Step
## Justification
Let \(h = g \circ f\). Then for any \(x \in X\),
\[
x \in h^{-1}(C)
\quad\Longleftrightarrow\quad
h(x) \in C
\quad\Longleftrightarrow\quad
g\bigl(f(x)\bigr) \in C.
\] But \(g\bigl(f(x)\bigr) \in C\) is the same as saying \(f(x) \in g^{-1}(C)\). Hence
\[
x \in h^{-1}(C)
\quad\Longleftrightarrow\quad
f(x) \in g^{-1}(C)
\quad\Longleftrightarrow\quad
x \in f^{-1}\bigl(g^{-1}(C)\bigr).
\] Because \(x\) was arbitrary, we conclude
\[
(g \circ f)^{-1}(C)
\;=\;
f^{-1}\bigl(g^{-1}(C)\bigr).
\] Thus the property is indeed **true** for all subsets \(C\) of \(Z\).
