1. functions

Prove or disprove
f:A-->B
g:C-->A
h:C-->A
if f o g = f o h, then g=h,

note: o denotes the composite function.
No idea on how to do this question.

2. Originally Posted by hugo84
Prove or disprove
f:A-->B
g:C-->A
h:C-->A
if f o g = f o h, then g=h,

note: o denotes the composite function.
No idea on how to do this question.
A function f is defined such that an element in the domain of f gets mapped to exactly one element in the codomain of f.

In other words, f(a) = f(b) if and only if a = b.

See how that applies?

3. Originally Posted by undefined
A function f is defined such that an element in the domain of f gets mapped to exactly one element in the codomain of f.
In other words, f(a) = f(b) if and only if a = b.
The statement in red is not true.
Consider this example: $\displaystyle A=\{a,b\},~B=\{1\},~\&~C\{x,y\}$
and $\displaystyle f=\{(a,1), (b,1)\},~g=\{(x,a),(y,b)\},~\&~h=\{(x,b),(y,a)\}$
Is it clear that $\displaystyle f \circ g = f \circ h$?
But is it true that $\displaystyle g=h$?

4. Originally Posted by Plato
The statement in red is not true.
Consider this example: $\displaystyle A=\{a,b\},~B=\{1\},~\&~C\{x,y\}$
and $\displaystyle f=\{(a,1), (b,1)\},~g=\{(x,a),(y,b)\},~\&~h=\{(x,b),(y,a)\}$
Is it clear that $\displaystyle f \circ g = f \circ h$?
But is it true that $\displaystyle g=h$?
Sorry, the "if" is true but not the "only if." Dumb mistake on my part. Thanks for correcting me.

Edit: Another very simple counterexample is:

f(x) = x^2
g(x) = x
h(x) = -x