- #1
Ed Quanta
- 297
- 0
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,
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,
Last edited: