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Ed Quanta
Apr20-04, 07:48 PM
I have to prove the following theorem,

1) If f:A->B is a surjection, and g:B->C is a surjection then g dot f:a->C is a surjection

Well this makes sense and I am not sure how to PROVE it

Is it sufficient to say the following

if for every element b of B, there exists a element a of A such that f(a)=b and same for B->C for g(b)=c is true, Then

for every element of c in C, there must exist an a of A such that g(f(a))=c since for every b in B there exists f(a)=b, and we know g(b)= c is true for every element of c,

matt grime
Apr21-04, 06:10 AM
Yes that's the right idea though the phrase 'we know g(b)=c is true for every..' is, erm, not what you want to write, but I think that is just the sentence structure nothing more.

You must show that for all c in C, there is an a in A such that gf(a)=c, which is what you've done.