# A one-to-one function in a surjective composition is bijective.

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.

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?
• 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. Oct 27, 2020 at 0:02