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How to prove the following

If $g\circ f$ is one-to-one and $f$ is onto, then $g$ is one-to-one.

Any assistance would be appreciated.

Daniel Sehn Colao
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user3753
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    https://math.stackexchange.com/questions/606257/if-g-circ-f-is-injective-and-f-is-surjective-then-g-is-injective – 19aksh Jun 01 '19 at 15:03

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Suppose $g(x)=g(y)$ for $x,y$ in the domain of $g$. As $g \circ f$ is defined, the domain of $f$ must equal the domain of $g$, and $f$ is onto, so there are $x',y'$ in the domain of $f$ such that $f(x')=x$ and $f(y')=y$.

Now it's easy to see that $(g \circ f)(x') = g(f(x'))=g(x)=g(y)= g(f(y'))=(g \circ f)(y')$ and so by injectivity of $g \circ f$ we conclude $x=y$ and we're done.

Henno Brandsma
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