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I have to prove that if a morphism h is coequalizer then it's an epimorphism. I'm trying to do this using the definition of coequalizer but I'm stuck. Any other idea?

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Let $f,g\colon X \to Y$ and $c\colon Y \to Z$ a coequalizer of $f$ and $g$. To prove that $c$ is epi, let $\alpha,\beta \colon Z \to A$ such that $\alpha c = \beta c$. Then we have $\alpha c f = \alpha c g$ and $\beta cf = \beta cg$. So $\alpha c$ and $\beta c\colon Y \to A$ coequalize $f$ and $g$. By the universal property of $c$, there is a unique $\gamma\colon Z \to A$ such that $\gamma c = \alpha c = \beta c$. As $\alpha$ and $\beta$ both have this property $\alpha = \beta$ and $c$ is epi.

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