# Prove that if $f \circ g = f\circ h$, then $g = h$

Given three functions,

$$f\colon A\to B, g\colon C\to A,h\colon C\to A$$

Prove or disprove that if $f \circ g = f \circ h$, then $g = h$.

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In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Nov 21 '12 at 16:47
It's certainly not true when $B$ is a singleton, unless $C$ is either a singleton of the empty set. –  Thomas Andrews Nov 21 '12 at 16:51

## 4 Answers

Here is a counterexample: \begin{align} g(1) & = 1 \\ g(2) & = 2 \\ g(3) & = 3 \\ \\ h(1) & = 1 \\ h(2) & = 3 \\ h(3) & = 2 \\ \\ f(1) & = 1 \\ f(2) & = 2 \\ f(3) & = 2 \end{align}

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+1 A counter example like this, tells the OP whole the story in a very fast way. –  Babak S. Nov 21 '12 at 16:54
g:C→A,h:C→A both gives A from a C input. So why does g(2) and h(2) produce different outputs, since they have the same C value? –  becozlah Nov 21 '12 at 17:15

A function with the property that $f \circ g = f \circ h \implies g = h$ would be one such that $f$ has a left inverse. This would mean $f$ is injective. Thus your statement holds if and only if $f$ is injective (or more generally $f$ is a monomorphism).

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Let $g:\mathbb R\to\mathbb R,g(x)=x$, $h:\mathbb R\to\mathbb R,h(x)=2x$, and $f:\mathbb R\to\mathbb R,f(x)=0$ for a counterexample.

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nice example, Won't Hunting! –  amWhy Nov 22 '12 at 0:01

Put $g(x) = x^2$, $h(x) =\cos(x)$ and $f(x) = 1$ for a real number $x$. We have $f\circ g = f\circ h$.

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