# 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.

## marked as duplicate by user228113, Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 23 '16 at 23:02

• Any thoughts on the question you might want to add? The question you've posed isn't very good as it stands. – Ali Caglayan Nov 23 '16 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
• 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. – tilper Nov 29 '16 at 2:52