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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. 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
up vote 6 down vote accepted

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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