Answer
Yes. If \(g \circ f\) is injective, then \(f\) must also be injective. There is **no** counterexample, because it is logically impossible for \(g \circ f\) to be injective unless \(f\) is injective.
Work Step by Step
## Proof (by Contradiction)
Assume \(g \circ f\) is injective, but suppose \(f\) is **not** injective. Then there exist \(x_1 \neq x_2\) in \(X\) such that
\[
f(x_1) \;=\; f(x_2).
\]
Apply \(g\) to these equal values:
\[
(g \circ f)(x_1)
\;=\;
g\bigl(f(x_1)\bigr)
\;=\;
g\bigl(f(x_2)\bigr)
\;=\;
(g \circ f)(x_2).
\]
Hence \(g \circ f\) takes the same value at two different points \(x_1\neq x_2\), which **contradicts** the assumption that \(g \circ f\) is injective.
Therefore, our assumption that \(f\) is not injective must be false. Thus, **\(f\) is injective**.
---
### Intuitive Explanation
If \(f\) “collapses” two distinct inputs \(x_1\neq x_2\) down to the same output in \(Y\), then after applying \(g\), those two points in \(X\) would still map to the same point in \(Z\). No further action by \(g\) can “undo” that collision. Hence, for the composition \(g \circ f\) to be injective, \(f\) itself must not collapse any distinct points—i.e., \(f\) must be injective.
