Answer
Yes. If \(g \circ f\) is onto \(Z\), then \(g\) itself must also be onto \(Z\).
Work Step by Step
## Proof
Suppose \(g \circ f\) is onto \(Z\). By definition, for **every** \(z \in Z\), there exists some \(x \in X\) such that
\[
(g \circ f)(x) \;=\; z.
\]
That means
\[
g\bigl(f(x)\bigr) \;=\; z.
\]
Set \(y = f(x)\). Since \(f(x)\in Y\), we have \(y \in Y\) and \(g(y) = z\). Hence, for **each** \(z \in Z\), we found a \(y \in Y\) with \(g(y) = z\). That is exactly the definition that \(g\) is onto \(Z\).
Therefore, whenever \(g \circ f\) is onto, \(g\) must also be onto.
