Here's the question:

Let $f: D \rightarrow Z$ be a function.

Show that: if function $g: Z \rightarrow D$ is given with charateristic:

$$\forall z \in Z: f(g(z)) = z$$

then $f$ is surjective.

My Proof: I just tried to prove this using definition of bijection of a function.

Firstly $\Rightarrow$:

It is already given that $\forall z \in Z: f(g(z)) = z$. So that means $g$ is bijective. That means that $\forall z \in Z, x \in D: g(z) = x$. In other words every $z$ in Pre-Image has atmost one image in $D$. So it is injective. Also $\forall x \in D$ there exists also a Pre-Image, which means its surjective.

Now $g$ is bijective. That means $f: D \rightarrow Z$ is inverse of $g$.

Therefore $f(x) = z$ and as $g(z) = x$ then $f(g(z) = z$ is obviously true.

So we can say that $f$ is sujective?

Or is there any concise way of proving it?

  • 3
    $\begingroup$ "It is already given that $\forall z \in Z: f(g(z)) = z$. So that means $g$ is bijective".This statement is false,this means only that $g$ is injective. $\endgroup$ Commented Nov 8, 2022 at 11:02
  • $\begingroup$ Then how come does $g$ have inverse $f$? $\endgroup$
    – viradia
    Commented Nov 8, 2022 at 22:56
  • $\begingroup$ What is truly strange about this problem: Here is a real and complete proof that $f$ is surjective: "$\forall z \in Z, f(g(z)) = z$, thus $f$ is surjective." $\endgroup$ Commented Nov 9, 2022 at 1:25

1 Answer 1


Alright I think I figured it out.

$f$ is bijective.

$g$ is bijective because


so that means there for every $x \in D$ there exists one and only one $z \in Z$ for function $f$. Other way round, for all $z \in Z$ there exists pre-Image in $x \in D$.

  • $\begingroup$ Consider: $Z = D = \Bbb R$, $$\begin{align}f(x) &= \begin{cases}0,& x =k\pi\text{ for some }k \in \Bbb Z\\\cot x,&\text{otherwise}\end{cases}\\g(x) &= \text{Arc}\,\text{cot } x\end{align}$$ $f(g(x)) = x$ for all $x \in \Bbb R$, but $f$ is not a bijection, and neither is $g$. $\endgroup$ Commented Nov 9, 2022 at 1:14

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