Question would be: prove/disprove that if $f\circ g$ injective and g is surjective, then f is injective.

after thinking, I came to the conclusion that it's a proof. tried to prove it but it looks not that valid. Would appreciate your feedback and corrections.


  • because $f\circ g$ is injective, then g is injective as well.
  • because it's given that g is surjective, and we came to conclusion it's also injective -> it's reversible by $g^{-1}$
  • if $f\circ g$ is injective and $g^{-1}$ is injective, then $f\circ g\circ g^{-1}$ injective as well.

Let there be $a_1,a_2$. $a_1=a_2 \iff f\circ g\circ g^{-1}(a_1)=f\circ g\circ g^{-1}(a_2) \iff f\circ i(a_1) = f\circ i(a_2) \iff f(a_1)=f(a_2)$

What do you think??

  • 2
    $\begingroup$ It is perfectly good. You could also consider working it without considering the reverse $g^{-1}$... $\endgroup$ – Clément Guérin Dec 27 '15 at 21:21

I do not think your proof is wrong per se, but I would go about things a little more directly.

Suppose $f(x_1)=f(x_2)$. There exist $y_1,y_2$ such that $x_1=g(y_1)$ and $x_2=g(y_2)$. We have $f\circ g (y_1)=f\circ g (y_2)$. So $y_1=y_2$. So $x_1=x_2$.


Let $f:B\to C$ and $g:A\to B$ and suppose $f\circ g$ is injective and $g$ is surjective. Then every element $b$ of $B$ is of the form $b=g(a)$, however you may not be able to define $g^{-1}$ since there may be multiple $a$'s with this property.

You should start with $f(b_1)=f(b_2)\implies f\circ g(a_1)=f\circ g(a_2)\implies a_1=a_2$ so that $b_1=b_2$, where $b_i=g(a_i)$. This is the more direct path instead of showing that $g$ is injective first so you can define $g^{-1}$.

  • $\begingroup$ He showed that $g$ is injective and surjective so there exists a two-sided inverse for $g$. $\endgroup$ – Future Dec 27 '15 at 21:23
  • $\begingroup$ But $g$ must be injective, for if it is not, then $f\circ g$ is not injective. $\endgroup$ – Michael Burr Dec 27 '15 at 21:24
  • $\begingroup$ Yeah. I saw that and fixed my answer. $\endgroup$ – Harry Reed Dec 27 '15 at 21:24

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.