How do I prove the following:
given $\sim$ on set X. Let $h:X \to (X/\sim)$. Also let $f:X \to Y$ be a function so that $x \sim y$ $\Longrightarrow$ $f(x)=f(y)$.
Prove that there is a $i:(X/\sim) \to Y$ so that $f=i \circ h$.
I have no clue how to even begin :(
P.S. group theory is not allowed.
I will assume that you mean (otherwise this doesn't make much sense):
(1) $\sim$ is an equivalence relation.
(2) $h$ sends a representative element of $X$ to its equivalence class: $h(x) = [x]$ where $[x] = \{ y \in X \;|\; x \sim y \}$.
Given an equivalence class, $[x]$, you need to define $i([x])$ so that $i(h(x))=i([x])$ equals $f(x)$. So $i([x])$ must be ???? (you have no choice).
Once you've made your definition, you need to worry about whether your definition depends on the choice of equivalence class representative. In particular, if $[x]=[y]$ (i.e. $x \sim y$), is it true that $i([x])=i([y])$? If not, your "function" $i$ isn't actually a well-defined function.
I hope this helps you get started! :)