I have been given the problem: "Say $$f:X→Y$$ and $$g:Y→Z$$are functions so that $$g◦f:X→Z$$ is bijective and $f$ is surjective. Can you deduce that $g$ is bijective?
My proof went as follows, I am not sure if this is valid?
$$ $$
Since $f$ is surjective then for any $y\in Y$ there exists and $x \in X$ such that $$f(x)=y$$
and $gf$ is bijective so it is surjective and injective so holds for:
$$g(f(x))=z$$for any $x$ and $$gf(x_1)=g(f(x_2)) \Rightarrow x_1=x_2$$
Then since we know $gf$ is surjective so $gf(x)=z$, since $f(x)=y$, then $g(y)=z$, so for any $z \in Z$ there exists $y \in Y$ such that $g(y)=z$, so $g$ is surjective.
Finally we say $$x_1=x_2$$ and call $h=gf$. Since $h$ is injective and surjective $$h(x_1)=h(x_2), g(f(x_1))=g(f(x_2))$$. Since we know that $f$ is surjective so $f(x)=y$ then $g(y_1)=g(y_2)$ so it is injective. Then since we have proven $g$ is both injective and surjective then it is bijective.
Best Answer
I don't believe you've proven injectivity of $g$. What you need to prove is that if for $y_1, y_2 \in Y$ you have $g(y_1) = g(y_2)$, then you have $y_1 = y_2$.
What you've proven is something weird starting from $x_1 = x_2$, and deduced $f(x_1) = y_1 = y_2 = f(x_2)$ and $g\circ f(x_1) = z_1 = z_2 = g \circ f(x_2)$ or something ? Which are obvious since both $f$ and $h$ are pure function, hence must send the same value to the same value regardless of their label in the latin or greek alphabet.
What I would do, though, is use the fact than bijective functions are also inversible. That is, there is a $h^{-1}$ such that $h \circ h^{-1}(z) = z$ and $h^{-1} \circ h (x) = x$ (prove it if you don't know that yet). Then if you have $g(y_1) = g(y_2)$ then you have $h^{-1}\circ g(y_1) = h^{-1}\circ g(y_2)$. I'll let you figure out the proper and complete proof.