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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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    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 $x\in X$.2011-12-09

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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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    Why do we $n$eed of the l$a$st requirement for the equivalence relation?2014-04-12
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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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The key point in building the coequalizer is to figure out the $"\sim"$ equivalence relation. This simply mean that we need to somehow partition the set Y.

Note that the smallest the equivalence relation, the largest the number of partitions. Thus, our goal is to partition Y as far as possible under certain conditions.

The largest number of partitions happens when we put each element of Y into one partition. However, we need to merge some of these partitions to satisfy the coequalizer definition. To this end, I think following the following pseudo code will will provide us with the desired partions:

$P = empty     // keep the partitions for each $y in Y{    build a new partition called $p;     put $y in $p;    put $p in $P;   } for each $x in X {         merge the partitions in $P that contain either f(x) or g(x);  }  

The above pseudo code build your partitions in Y. This might be the one that you can get with the smallest equivalence relation with the conditions explained for the coequalizer.