If g and f o g are onto(Surjective), is f onto(Surjective)?

[LittleTexan]

Hello Members

I am having a little bit of a problem solving this proof for my Discreet math course.

If g and f o g are onto(Surjective), is f onto(Surjective)? Need to prove. I believe that f has to be Onto.

so I have g: A -> B
f: B -> C

Well I understand that a function is onto(Surjective) when it maps to all images. So for g all the elements in A map(hit) element in B.

So f o g: A -> C where every element of C must be map to. I believe that since C is the Image of f that this means that f must be onto(Surjective).

Can someone give me advise on how to prove this? Or even just some advise in general on proofs.

TIA

 

[matt grime]

The image of fg is a subset of the image of f. If f is not surjective then fg cannot possibly be surjective.