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Suppose that $g \circ f$ is the composition of $f:X \to Y$ and $g:Y \to Z$. Suppose that $g \circ f$ is onto.

  1. Prove that if $g$ is one-to-one, then $g$ is bijective and $f$ is onto.

This is the last part of a series of problems stemming from the original statement; I proved that g is surjective, but this part of the problem has me stumped. Maybe it's something really obvious that I'm missing, but I generally don't have a good grasp on injections, surjections, and bijections, so maybe not.

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Since you have already proved $g$ is onto and it is given that $g$ is one-one, $g$ must be bijective.

Suppose $f$ is not onto. Then $\exists y\in Y|y\notin f(X)$. Thus $g(y)\notin g(f(X))$ violating the surjectivity of $g\circ f$. Do you notice how we used the one-oneness of $g$ in this conclusion?

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    $\begingroup$ Why argue by contradiction? Since $g$ is bijective, the composition of $g^{-1}$ and $g\circ f$ is a composition of surjective functions, hence surjective. $\endgroup$
    – Arturo Magidin
    Oct 27, 2020 at 0:02
  • $\begingroup$ Sure, that would be a good way too! $\endgroup$
    – Shubham Johri
    Oct 27, 2020 at 0:04
  • $\begingroup$ Yes, that makes sense. Thanks $\endgroup$
    – MC5555
    Oct 27, 2020 at 11:31

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