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

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SUMMARY

If functions g: A -> B and f o g: A -> C are both onto (surjective), then function f: B -> C must also be onto (surjective). This conclusion is based on the definition of surjectivity, where every element in the codomain must be mapped by at least one element from the domain. Since f o g maps all elements of A to C, and g maps all elements of A to B, it follows that f must cover all elements in C to maintain the surjectivity of the composite function f o g.

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

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