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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.

Proof:

  • 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??

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    $\begingroup$ 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
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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 $x_1=x_2$.

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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 $g^{-1}$.

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  • $\begingroup$ 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

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