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I have two questions which are related to mappings as follow:

Given 3 sets E, F, and G such that $f: E \rightarrow F, g: F \rightarrow G$ are two mappings. Prove that:

a. If $g \circ f$ is injective and $f$ is surjective then $g$ is injective.

b. If $g \circ f$ is surjective and $g$ is injective then $f$ is surjective.

For these two problems. I have solved halfway for each.

For the problem (a), I have proved that $f$ is bijective.

For the problem (b), I have proved that $g$ is bijective.

I don't know how to do next because I don't find the relation to conclude.

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    $\begingroup$ Do you know that the composition injective functions is injective, and the composition of surjective functions is surjective? $\endgroup$
    – user804886
    Jul 12, 2020 at 2:37
  • $\begingroup$ @user804886 Oh I see, is that the next step to solve? $\endgroup$
    – Hoang Nam
    Jul 12, 2020 at 2:37
  • $\begingroup$ I can see one way of finishing the proof using these facts, yes. $\endgroup$
    – user804886
    Jul 12, 2020 at 2:38
  • $\begingroup$ @user804886 Thank you, I think it is right to solve this $\endgroup$
    – Hoang Nam
    Jul 12, 2020 at 2:39

1 Answer 1

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In part a, because $g \circ f$ is injective and $g^{-1}$ (which exists, since $g$ is bijective) is injective, we therefore have $$f = g^{-1} \circ (g \circ f)$$ is the composition of injective functions, which hence makes it injective. Part b can be solved dually.

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