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Injective and Surjective Functions

If $g(f(x))$ is one-to-one (injective) show $f(x)$ is also one-to-one given that $f$ is a function from $A$ to $B$ and $g$ a function from $B$ to $C$.

I've just started my Discrete math course and I'd like some help on this. I'm pretty sure we're supposed to use set theory laws to prove this.

So far I know the three conditions that satisfy an injective function (sorry, having difficulties typing all this TeX markup so I'll skip that).

Any help?

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    This is not really discrete mathematics, as discreteness is not used in the proof.2011-03-20

3 Answers 3

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Hence $g \circ f$ is injective then by definition it means that for all $x, y$ in the domain of $f$ (being the domain of $g \circ f$) we get $[(g \circ f) (x) = (g \circ f)(y)] \Rightarrow x = y.$ Assume now that $f(x) = f(y)$. Then of course $(g \circ f)(x) = g(f(x)) = g(f(y)) = (g \circ f)(y)$. By previous implication we obtain $x = y$ and hence $f$ is injective.

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Or the other way around: Take $x \ne y$. Since $g \circ f$ is injective we have $g(f(x))\ne g(f(y))$, so we must have $f(x)\ne f(y)$.

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    @AnthonyPeter: No, this is just saying that that if $g(a)\neq g(b)$ then we must have $a\neq b$.2014-11-10
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Hint: suppose that $f$ is not one to one. Then there are $x \ne y$ such that $f(x)=f(y)$. Can you conculude that $g\circ f$ is not one to one?

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    anon is suggesting that you argue by contraposition, in other words show that if f is not injective then g(f) isn't either. If you want to show g(f) isn't injective you need to find two distinct points in A that g(f) sends to the same place. Anons comment will help you do that.2011-03-20