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This question already has an answer here:

Note :This may be similar to some questions but its not the same, i checked.

The question is : Decide if the following statement is true or false and prove your claim :

If $ f \colon A \to B \text{ and } g \colon B \to C $ such that $ g \circ f$ is surjective, then $f$ is surjective.

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marked as duplicate by user228113, Asaf Karagila elementary-set-theory Nov 23 '16 at 23:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Any thoughts on the question you might want to add? The question you've posed isn't very good as it stands. $\endgroup$ – Ali Caglayan Nov 23 '16 at 22:42
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Lets assume $g:\{0,1\} \to \{0\}: 0 \to 0, 1 \to 0$ and $f: \{1,2,3\} \to \{0,1\}: 1 \to 0, 2 \to 0, 3 \to 0$ than $ g \circ f$ is surjective, but not f

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Assume if g o f is surjective then f is surjective . But for arbitrary f: A>B consider g:B>ran(f) which is the identity over the range of f. g o f is surjective so f is always surjective onto B. This is absurd.

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  • $\begingroup$ I edited it to make it a good proof. Please remove downvotes. $\endgroup$ – Jacob Wakem Nov 23 '16 at 23:15
  • $\begingroup$ A little clearer now but in general it's better to just provide one explicit counterexample if all you want to do is show something of the form "$P \implies Q$" is false. $\endgroup$ – tilper Nov 29 '16 at 2:52
  • $\begingroup$ @tilper no its not. its not just about the seconds at the end. $\endgroup$ – Jacob Wakem Nov 29 '16 at 3:08
  • $\begingroup$ I don't know what "it's not just about the seconds at the end" means but the standard way of showing a statement isn't true is to give an explicit counterexample. $\endgroup$ – tilper Nov 29 '16 at 4:03
  • $\begingroup$ @tilper Oh I know its the standard way. $\endgroup$ – Jacob Wakem Nov 29 '16 at 5:51

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