> [!NOTE] **Theorem** (Universal Property of Quotients) > Suppose that $X$ and $Y$ are sets and that $f:X \to Y$ is a [[Well-defined Function with Respect to Equivalence Relation|well defined]]. There is exactly one function $f_{E}: X/E\to Y$ so that $f=f_{E}\circ q_{E}$ where $q_{E}:X\to X/E$ defined by $q_{E}(x)=[x]_{E}$ is the [[Quotient Map|quotient map]] given by $E$. **Proof.** Let $Q$ be the relation from $X/E$ to $Y$ which contains $([x]_{E},y)$ exactly when there is some $x' \in X$ so that $f(x')=y$ and $x' \in[x]_{E}$. We now must show that $Q$ is graphical. Fix $[x]_{E}$. Because $E$ is reflexive, $x$ lies in $[x]_{E}$. Let $y=f(x)$. So $([x]_{E},y)$ lies in $Q$. Suppose now that there are $y',y''\in Y$ so that $([x]_{E},y')$ and $([x]_{E},y'')$ lie in $Q$. By the defining property of $Q$ there are $x'$ and $x''$ so that $f(x')=y'$ and $f(x'')=y''$. By the definition of $[x]_{E}$ we have $xEx'$ and $xEx''$. By symmetry of $E$ we have $x'Ex$. By transitivity of $E$ we have $x'Ex''$. Since $f$ is well defined $f(x')=f(x'')$ and so $y' = y''$. Thus $Q$ is graphical. Taking $f_{E}=Q$ completes the construction of $f_{E}$.