quotient of an object 2

Question

Here I've asked what is a quotient of an object and the answer was that it is an equivalence class of epis.

But here on the first page they claim that regular quotient is an coequalizer of 2 morphisms. I do not follow why this coequalizer should be epi.

Without compiling the postscript, are you sure the paper didn't mean coequaliser? — user2628206, Sep 26 '21 at 12:47
@user2628206 A mouseslip, you are right. So what is the next step to answer my question ? — user122424, Sep 26 '21 at 15:00

Alex Kruckman · Accepted Answer · 2021-09-26T16:41:24.627

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It seems that your question is: If $c\colon B\to C$ is the coequalizer of the pair $f,g\colon A\to B$, why is $c$ an epimorphism?

To show that $c$ is an epimorphism, suppose $p,q\colon C\to D$ are two arrows such that $p\circ c = q\circ c$. We need to show that $p = q$.

Let $h = p\circ c = q\circ c\colon B\to D$. Since $c$ coequalizes $f$ and $g$, we have $c\circ f = c\circ g$, so $h\circ f = p\circ c \circ f = p\circ c \circ g = h\circ g$. Thus $h$ coequalizes $f$ and $g$. By the universal property of the coequalizer, there is a unique arrow $h'\colon C\to D$ such that $h'\circ c = h$. By uniqueness, $h' = p = q$.

An arrow is called a regular epimorphism if it is a coequalizer for some parallel pair of arrows. Dually, every equalizer is a monomorphism, and an arrow is called a regular monomorphism if it is an equalizer for some parallel pair of arrows.

edited Sep 26 '21 at 16:41

answered Sep 26 '21 at 15:13

Alex Kruckman

76,357

There is a shorter proof here where I do not understand : By the universal property of $c$, there is a unique $\gamma\colon Z \to A$ such that $\gamma c = \alpha c = \beta c$ – user122424 Sep 26 '21 at 17:13
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@user122424 But the proof you link to is exactly the same argument I gave, just using different variable names... – Alex Kruckman Sep 26 '21 at 17:15
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Also, it's not meaningfully shorter. If you delete the paragraph break and the explanatory "We need to show that p = q" in my proof, then both proofs are exactly four lines long! – Alex Kruckman Sep 26 '21 at 17:18
I understand your proof but I have a difficulty to translate your proof into their : why$\gamma c = \alpha c = \beta c$. – user122424 Sep 26 '21 at 17:44
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Their $\alpha$ is my $p$. Their $\beta$ is my $q$. Their $\gamma$ is my $h'$. Their equation $\gamma c = \alpha c = \beta c$ is my $h'\circ c = h$ (since $h = p\circ c = q\circ c$). – Alex Kruckman Sep 26 '21 at 17:46

quotient of an object 2

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