I have a math question from computer science. The following should be a fundamental fact from mathematics. Can you the mathematicins tell me how you would say it in a more elegant way?
Given
- a mapping f: A=>B
- an equivalence relation ~ on A
satisfying,
for each a1,a2 of A, a1 ~ a2 implies f(a1)=f(a2)
Then, there exists a unique factorisation of the mapping f
$f= g \circ h$
such that
- h is a mapping from A to a A/~
- g is a mapping from A/~ to B.
There must be some more elegant way to say this in mathematics, like some algebra thing with isomorphism, congruence, or quotient etc. as keywords. I would like to avoid terms from category theory if possible, because that is too much for most computer scientists.
So, I am looking for a mathematical way to say the above fact. It seems really close to the first isomorphism theorem. Any idea? Thanks.
Without loss of generality let $f:A\rightarrow B$ is onto [if not take $B=range(f)$] then from the quotient set of $A$ w.r.t given equivalence relation the induced function becomes $1-1$ (from your observation it is $g$).
You already observed the existence. Only remaining part is this decomposition is unique. Since your $h$ is unique (canonical map) and $g$ is invertible there is no other way to factorize $f$.
Q.E.D