Answer
False.** One‐to‐one (\(\mathrm{injective}\)) functions \(h\) need not force \(f\) and \(g\) to agree on all of \(X\); they only force agreement on the *image* of \(h\). If that image is not all of \(X\) (which can happen in infinite sets), \(f\) and \(g\) can differ outside that image yet still satisfy \(f\circ h = g\circ h\).
Work Step by Step
## A Concrete Counterexample
Take \(X = \mathbb{N} = \{0,1,2,3,\dots\}\). Define:
1. **Injective** \(h : \mathbb{N} \to \mathbb{N}\) by
\[
h(n) = n+1.
\]
This is clearly one‐to‐one. But it is **not** onto, since 0 is never in its range.
2. Two functions \(f,g : \mathbb{N}\to \mathbb{N}\) that agree on all inputs \(\ge1\) but differ at 0. For instance:
\(g(x) = 0\) for **all** \(x\in\mathbb{N}\).
\(f(x) =
\begin{cases}
1, & x=0,\\
0, & x\ge1.
\end{cases}
\)
### Check \(f \circ h = g \circ h\)
For any \(n\in\mathbb{N}\),
\(\;h(n) = n+1 \ge 1.\)
Hence \(f\bigl(h(n)\bigr)=f(n+1)=0\) and \(g\bigl(h(n)\bigr)=g(n+1)=0.\)
So \(\;(f\circ h)(n)=(g\circ h)(n)=0\) for **every** \(n\). Therefore,
\[
f\circ h \;=\; g\circ h.
\]
### But \(f\neq g\)
We see \(f(0)=1\) whereas \(g(0)=0\). Thus \(f\) and \(g\) are genuinely different as functions on \(X\).
---
## Why This Happens
- Composition \(f\circ h\) only “tests” \(f\) on the set \(\mathrm{Im}(h)\subset X\).
- If \(h\) is **not** onto, there can be points in \(X\) (like \(0\) here) that are **never** hit by \(h\). On those points, \(f\) and \(g\) can disagree without affecting \(f\circ h\) or \(g\circ h\).
- Consequently, \(f\circ h = g\circ h\) does **not** force \(f=g\).
Hence the statement
> “If \(h\) is injective and \(f\circ h = g\circ h\), then \(f=g\)”
is **false** in general. It *would* be true if \(h\) were **onto** (surjective) instead.
