# Co-equalizers in Set

http://mathworld.wolfram.com/Coequalizer.html

This link indicates how to form co-equalizer in category of Set. I have been given homework to describe co-equalizer for a variant of Set category but I am unable to understand completely how to form it in Set. Can someone please explain in "easy way" how to form co-equalizers in category Set?

Thank you.

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Can you say which part of the explanation confuses you? –  Zhen Lin Dec 9 '11 at 21:12
What is minimal equivalence relation on Y that makes $f$ and $g$ equal? More precisely, what is the canonical map $c$ in this case? –  user18096 Dec 9 '11 at 21:32
The minimal equivalence relation is... the minimal equivalence relation. If you don't understand that then the problem is not with category theory. The canonical map is simply the quotient map taking an element $x$ to its equivalence class $[x]$. –  Zhen Lin Dec 9 '11 at 21:36
I meant how to define the equivalence relation ~ here. Is it right to define $y_1$ ~ $y_2$ when $f(x)=g(x)$ for all $x\in X$? Is it minimal? –  user18096 Dec 9 '11 at 21:46
What you write makes no sense. $y_1\sim y_2$ has no letter in common with $f(x)=g(x)$. You start with a relation that makes $f(x)\sim g(x)$ for each $x\in X$. Chances are, this by itself will not be an equivalence relation, so you consider the smallest equivalence relation that contains all pairs of the form $(f(x),g(x))$ (the intersection of all equivalence relations that contain all such pairs). This is an equivalence relation, and its the smallest equivalence relation in which $f(x)\sim g(x)$ holds for each $x\in X$. –  Arturo Magidin Dec 9 '11 at 21:53

Let $X$ and $Y$ be sets, and let $f, g : X \to Y$ be maps. The coequaliser of $f$ and $g$ is the quotient map $c : Y \to Y / \sim$, where $\sim$ is the smallest equivalence relation such that $f(x) \sim g(x)$ for all $x$ in $X$. So, for example, the following will be true:

• For all $y$ in $Y$, $y \sim y$.
• For all $x_1, \ldots, x_n$ in $X$ such that $g(x_1) = f(x_2), g(x_2) = f(x_3), \cdots, g(x_{n-1}) = f(x_n)$, we have $f(x_1) \sim g(x_n)$.
• If $g(x_1) = g(x_2)$ then $f(x_1) \sim f(x_2)$.

Working out what the equivalence relation is explicitly is not very enlightening; it is far more useful to understand the universal property of a coequaliser.

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thanks very helpful. –  user18096 Dec 9 '11 at 22:08
I would say that not working out explicitly the equivalence relation will only result in the result seeming magical. In particular, if person X cannot tell me when to elements of the equalizer are equal in terms of the initial data of the problem, it is rather hard to conclude that she has any idea of what the coequalizer is. –  Mariano Suárez-Alvarez Dec 10 '11 at 3:13
Why do we need of the last requirement for the equivalence relation? –  user42912 Apr 12 '14 at 13:43

One particular example of coequalizer may be known to you. Let's say we are in the category of vector spaces, or abelian groups. Let $f: V \longrightarrow W$ be any linear map (group morphism) and $0: V \longrightarrow W$ the zero map (morphism), that is, the map which sends every element of $V$ to zero. Then, the coequalizer of $f$ and $0$ is just the cokernel of $f$, $\mathrm{cok}\ f = W / \mathrm{im}\ f$.

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