TITLE: Coequalizer in $\mathsf{Sets}$
QUESTION [6 upvotes]: Assume $f,g:X \to Y$ are arrows in $\mathsf{Sets}$. Then the coequalizer is given by $c:Y \rightarrow Y/R$ where $R \subseteq Y\times Y$ is the smalles equivalence relation on $Y$ s.t. $\forall x \in X: (f(x),g(x)) \in R$. Given any $h:Y \rightarrow Z$ there is a unique $\overline h: Y/R \to Z$ s.t. $\overline h \circ c = h$. I know that $\overline h ([y]) = h(y)$. 
My question: How can I prove that $[y]=[y'] \Rightarrow h(y) = h(y')$ ?

REPLY [2 votes]: Given any $h:Y\to Z$ such that $hf=hg$ ...
Now, "have the same image under $h$" determines an equivalence relation $H$, and since $(f(x),g(x))\in H$ it follows that $R\subseteq H$.