Question

7. (New) Let g : A-> B be a function and let f : B-+ C be an injection. If f° g : A surjective, then g is surjective C is
write the proof

Answer 1

$g:A\to B$ is some function and $f:B\to C$ is injection

If $f\circ g:A\to C$ is surjective then we want to prove that $g:A\to B$ must be surjective

That is, we want to show that for every $b\in B$, there exists $a\in A$ such that

$g(a)=b$

As $f:B\to C$ is a function, we have $f(b)=c\in C$ exists

And so we have $f\circ g:A\to C$ is a surjective function means $\exists a\in A$ such that

$(f\circ g)(a)=c\in C$

That is, $f(g(a))=c$

But we already have $f(b)=c$

So that $f(g(a))=f(b)\Rightarrow g(a)=b$ via injectivitiy of $f:B\to C$

And so $\exists a\in A$ such that $g(a)=b$

Which means $g:A\to B$ must be surjective