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Suppose a composition $g\circ f$ that is injective. Prove $f$ is too.

Where $f:A\mapsto B$ and $g:B\mapsto C$

If $g\circ f$ is injective then $(g\circ f)(a_1)=(g\circ f)(a_2)\implies f(a_1)=f(a_2)$, which proves that $g$ is injective.

But I have a problem trying to prove $f$ is injective.

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    It's not necessarily true that $g$ is injective if $g\circ f$ is injective by the way. And for the proof try contradiction?? – 2017-02-28
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    Assume $f$ is not injective and immediately get a contraddiction. – 2017-02-28

2 Answers 2

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Assume $f(x) = f(y)$. Then apply $g$, you get $g(f(x)) = g(f(y))$. Since $g\circ f$ is injective, you get $x=y$.

Hence we proved that $f(x)=f(y)$ implies $x=y$ which is the definition of injectivity.

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Hint:

Use contrapositive:

If $f$ is not injective, $g\circ f$ can't be injective.