If $g\circ f$ is surjective, then $f$ is surjective. [duplicate]

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.

• Any thoughts on the question you might want to add? The question you've posed isn't very good as it stands. Commented Nov 23, 2016 at 22:42

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

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.

• I edited it to make it a good proof. Please remove downvotes. Commented Nov 23, 2016 at 23:15
• 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.
– user307169
Commented Nov 29, 2016 at 2:52
• @tilper no its not. its not just about the seconds at the end. Commented Nov 29, 2016 at 3:08
• 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.
– user307169
Commented Nov 29, 2016 at 4:03
• @tilper Oh I know its the standard way. Commented Nov 29, 2016 at 5:51