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.

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.