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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 $[y]=[y'] \Rightarrow h(y) = h(y')$ ?

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What is $c$? You have not defined it. And if it is a quotient map then $h$ would be also forced to be equal on objects of the same equivalence class. I misunderstand something –  porton Oct 26 '13 at 20:18
    
@porton: He forgot to mention that $hf$ must be the same as $hg$, see my answer below. –  Stefan Hamcke Oct 27 '13 at 22:35

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up vote 2 down vote accepted

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 $R\subseteq H$.

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For clarity: Make $H \subseteq Y^2$ equivalence relation s.t. $(y,y') \in H \iff h(y) = h(y')$. Then since $hf = hg$ we have $\forall x \in X: (f(x),g(x)) \in H$. With the minimality of $R$ we know $R \subseteq H$. Then if $[y]=[y']$ we have $(y,y') \in R$ and thus $(y,y') \in H$ s.t $h(y) = h(y')$. Right ? –  André Oct 10 '13 at 15:58
    
@André: Yes, that's exactly how it works. –  Stefan Hamcke Oct 10 '13 at 16:01
    
Thanks. It seemed strange to me that those minimal equivalence relations exist. But now I recognize that the intersection of eq. relations is again an equivalence relation and $\mathcal P(Y^2)$ is always one. –  André Oct 10 '13 at 16:06
    
@André: Not $\mathcal P(Y^2)$ but $Y^2$. –  Stefan Hamcke Oct 10 '13 at 16:15
    
Oops yes, sure. –  André Oct 10 '13 at 19:06

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