Answer
**No**, \(f\) need not be onto even if \(g \circ f\) is onto.
Work Step by Step
## Counterexample
Let’s define three sets and two functions:
- **Sets**:
- \(X = \{\,1,2\}\)
- \(Y = \{\,a,b,c\}\)
- \(Z = \{\,p,q\}\)
- **Function** \(f : X \to Y\):
\[
f(1) = a,\quad f(2) = b.
\]
Notice \(f\)’s image is \(\{a,b\}\subset Y\). Since \(c\) is never used, \(f\) is **not** onto \(Y\).
- **Function** \(g : Y \to Z\):
\[
g(a) = p,\quad g(b) = q,\quad g(c) = p \;(\text{for example}).
\]
Observe that \(\{a,b\}\) (the image of \(f\)) maps onto all of \(Z\). Specifically, \(g(a)=p\) and \(g(b)=q\).
### Check that \(g \circ f\) is onto \(Z\)
1. \((g \circ f)(1) = g\bigl(f(1)\bigr) = g(a) = p.\)
2. \((g \circ f)(2) = g\bigl(f(2)\bigr) = g(b) = q.\)
Hence the range of \((g \circ f)\) is \(\{p,q\}\), which is exactly all of \(Z\). Therefore, **\(g \circ f\) is onto** \(Z\).
### But \(f\) is Not Onto \(Y\)
We see \(f\) never takes the value \(c\). Thus \(\operatorname{Im}(f)=\{a,b\}\neq Y\). So \(f\) is **not** onto \(Y\).
---
## Conclusion
This example proves that even if \((g \circ f)\) is onto \(Z\), the intermediate function \(f\) itself need **not** be onto \(Y\).
