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Let $f:A\to B$ and $g:B\to C$ be the two functions ?

Then, if $g\circ f:A\to C$ is onto, then what can be said about $g:B\to C$ ?

My try :

I have $g\circ f:A\to C$ here, so I can say that

" Every element $z\in C$ there exists $y\in A$ such that $g\circ f(y) = z$ "

This gives $g(f(y)) = z$

Now, this shows a form which is implying that $g$ is onto.

Am I right here ?

  • 1
    Yes, and that is all you can say. Although the proof is somewhat too hand-wavy for me. – 2017-01-13
  • 3
    I have two things to say: `\to` and `\circ`. – 2017-01-13

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