# Composition function

If $f:S \to T$ and $g:T\to U$ are functions how can I prove that if $(g o f )$ is one-to-one, so is $f$, and find an example where $(g o f )$ is one-to-one but $g$ is not one-to-one.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. 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. –  Julian Kuelshammer Nov 4 '12 at 15:37
@JulianKuelshammer thank you !=] –  user1798157 Nov 4 '12 at 16:04

I'll try to give you some general hints instead of the solution, since I think that will be more helpful for your future life with mathematics:

Try prove by contradiction: Suppose $f$ was not one-to-one. What does this mean? Then, how is $g\circ f$ defined?

To construct an example where $g\circ f$ is one-to-one, but $g$ is not, write down some functions that you know of, that are not one-to-one. Sometimes it helps to restrict the domain or similar things to chose a suitable $f$ for such $g$.

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HINT for the example:

$\qquad\qquad$

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