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.
 A: 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.
A: Although a correct answer has already been given, I’ve just realized that there is somewhat more structural and categorical way to represent co-equalizers in $Set$.
A co-equalizer is a co-limit of the diagram: $A \rightrightarrows D$.
Recall, that diagrams $A \rightrightarrows D$ in $Set$ are just graphs. Strictly speaking, they form a category of directed irreflexive graphs $Set^{\circ \rightrightarrows \circ}$ but since we do not distinguish between $f: A \to D$ and $g: A \to D$, we can treat arrows $a \in A$ just as “edges” between dots $d \in D$.
It is clear that for any co-equalizing $e^{\prime}$ $A \overset{f; e^{\prime}}{\underset{g; e^{\prime}} \rightrightarrows} E^{\prime} $ is also a graph. Moreover, since $f;e^{\prime} = g;e^{\prime} $ this graph is discrete (contains only loops). Also $e^{\prime}$ defines a morphism in the category of graphs since it clearly doesn’t break arrows.
Now, our task is to find the finest epic morphism $e$ (co-equalizers are always epic) from the graph $A \overset{f}{\underset{g} \rightrightarrows} D$ to some discrete graph $A \overset{f;e}{\underset{g;e} \rightrightarrows} E$. The most natural way is to simply factorize dots of $D$ by a relation of “being connected by arrows from $A$”. Which gives the desired construction.

I’ve found quite enlightening to think about diagrams of form $A \overset{f}{\underset{g} \rightrightarrows} E$ in any category as a “connection of two ways of comparing $A$ and $D$ by means of $A$” and co-equalizer as a minimal way to “collapse” $D$ over this connection.
A: 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.
A: You can get a description of how to calculate coequalisers on finite sets from the book Computational Category Theory by David Rydeheard and Rod Burstall. (Acknowledgements also to D. T. Sannella and others.) It's available at http://www.cs.man.ac.uk/~david/categories/ , together with code for the programs therein contained. The algorithm for coequalisers is on page 102 of the PDF.
Here are two simple coequalisers, with the one on the right partitioning by coalescing two elements. This isn't the only way to think about coequalisers: as Zhen Lin says, it's good to think about their universality. But it's also good to see details. For example, if you calculate several coequalisers on the same sets but with different parallel arrows you get a feel for how the size of the equivalence relation varies with the number of partitions, as pointed out by qartal. 
A: 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 $f$, $\mathrm{cok}\ f = W / \mathrm{im}\ f$.
