Thread: urgent homework-proving surjectivity from conditions

I am arguing from the given supposition:
if g(f(x))=h(f(x)) then g(y)=h(y)
(f maps from A to B, g and h map from B to C)

to the conclusion that f is surjective

I'm trying to argue to a contradiction by supposing that f isn't surjective...do you think that will work? Any suggestions?

Originally Posted by blackmustachio

I am arguing from the given supposition:
if g(f(x))=h(f(x)) then g(y)=h(y)
(f maps from A to B, g and h map from B to C)

to the conclusion that f is surjective

I'm trying to argue to a contradiction by supposing that f isn't surjective...do you think that will work? Any suggestions?

Say for all and specific . I constructed a function from to which collapses all its elements into a single element. Now is surly not surjective (unless has a single element, so say has more than just . Why?) And let be the exact same function from to . So we have for all , and we also have for all by construction. But is not surjective. Meaning what you are trying to prove is completely wrong.

Let C = {0,1}.

Assume f is not surjective (toward contradiction). Thus there is some b in B such that b is not in the range of f.

Define g:B to C by g(x) = 0.

Define h:B to C by h(b)= 1, and h(x) = 0 whenever x is not b.

g(f(x)) = h(f(x)) = 0 for all x in A.

but g(b) = 0 and h(b) =1, thus h is not g.