# urgent homework-proving surjectivity from conditions

• Sep 20th 2007, 11:30 PM
blackmustachio
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?
• Sep 21st 2007, 08:16 AM
ThePerfectHacker
Quote:

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?

$f: A\mapsto B$
$g: B\mapsto C$
$h: B\mapsto C$

Say $f(x) = b$ for all $x\in A$ and specific $b\in B$. I constructed a function from $A$ to $B$ which collapses all its elements into a single element. Now $f$ is surly not surjective (unless $B$ has a single element, so say $B$ has more than just $b$. Why?) And let $g,h$ be the exact same function from $B$ to $C$. So we have $g(f(x)) = h(f(x))$ for all $x\in A$, and we also have $g(y) = h(y)$ for all $y\in B$ by construction. But $f$ is not surjective. Meaning what you are trying to prove is completely wrong.
• Sep 11th 2008, 12:26 PM
quote7000
ddddddddd
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.