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Suppose $e: B \rightarrow C$ is the coequalizer of two parallel morphisms $f,g; A \rightarrow B$.

Show that if $e$ is monic then it is an isomorphism.

I know that $e$ is epic.

If $e$ is monic then there exists $\alpha \text{ } \beta : X \rightarrow B$ such that if $e \circ \alpha = e \circ \beta$ then $\alpha = \beta$

To show $e$ is an isomorphism, do I have to show that there exists $f: C \rightarrow B$ such that $e \circ f(c) = c$ and that $ f \circ e(b) = b$?

How do I go about proving this?

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2 Answers 2

From $e \circ f= e \circ g$ and the monic hypothesis you get $f=g$, then as $1 \circ f= 1 \circ g$ you get a factorization $h$ such that $h \circ e=1$. You conclude since your $e$ is both a section and an epimorphism, thus it's an isomorphism.

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I recommend doing this in several steps:

  1. Show that a coequaliser is always epic. A coequaliser is also called a regular epimorphism (actually a coequaliser is not just the arrow $e$ itself but rather the object $C$ together with a family of maps from the diagram $A\stackrel{\longrightarrow}\longrightarrow B$ to $C$, but you will learn what that means when you learn about limits.)
  2. Show that a regular epimorphism is an extremal epimorphism. An epimorphism is called extremal if for every factorization $e=mf$ where $m$ is monic, $m$ is also an isomorphism.
  3. Show that an extremal epic which is also monic is an isomorphism.
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