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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?

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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.

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    $\begingroup$ you can use single dollar symbols $ to enclose inline math, like this $z \in Z$ : $z\in Z$. Please update formatting. $\endgroup$ – Lærne Mar 19 at 20:44
  • $\begingroup$ Will change now thank you $\endgroup$ – Olly Reynolds Mar 19 at 20:47
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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.

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  • $\begingroup$ Thank you, i will look at it now! $\endgroup$ – Olly Reynolds Mar 19 at 21:02
  • $\begingroup$ How did you get $h^{-1}\circ g(y_1) = h^{-1}\circ g(y_2)$? $\endgroup$ – Olly Reynolds Mar 19 at 21:06
  • $\begingroup$ would you mind completing the proof? $\endgroup$ – Olly Reynolds Mar 19 at 21:44
  • $\begingroup$ Hum... if $z_1 = z_2$, then obviously $h^{-1}(z_1) = h^{-1}(z_1)$ else $h^{-1}$ would be a ill-defined function. If the values are the same, then the function must send them to the same image value ! In this case we have $z_1$ to be $g(y_1)$... and likewise $z_2$ is $g(y_2)$. I still believe you can complete the proof on your own. $\endgroup$ – Lærne Mar 20 at 15:18
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You can prove it more easily(I think much more) by using RAA argument. That is, first suppose that gf is not bijective and lead a contradiction from the supposition and the fact that f is a "surjective" "function".

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