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.


1 Answer 1


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?

  • 1
    $\begingroup$ 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. $\endgroup$ Oct 27, 2020 at 0:02
  • $\begingroup$ Sure, that would be a good way too! $\endgroup$ Oct 27, 2020 at 0:04
  • $\begingroup$ Yes, that makes sense. Thanks $\endgroup$
    – MC5555
    Oct 27, 2020 at 11:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.