Picture - Equivalence Relation & Classes, Partitions, Quotient Set, & other related ideas To get intuition for them and to remember them, I'd be grateful for a picture that combines and embodies the key definitions regarding Equivalence Relations & Classes, Quotient Sets, and Partitions. 
MSE suggests https://math.stackexchange.com/q/641452/53259 as a related question, but is there a better picture (I don't understand the reason for its closure)? 
 A: I hope this helps. I like to think of equivalence relations as defining a new 'type' of equality. Remember an equivalence relation must be reflexive ($a$~$a$), symmetric (if $a$~$b$ then $b$~$a$) and transitive (if $a$~$b$ and $b$~$c$, then $a$~$c$).
For example if you want to work in $\mathbb{Z}/5\mathbb{Z}$ which is like the integers modulo $n$, you place a equivalence relation of $\mathbb{Z}$ by saying that $a$~$b$ if and only if $a-b=5n$ for some $n\in\mathbb{Z}$. This just says that if $a$ and $b$ differ by a multiple of $5$, then they are considered 'equal', or to be more precise equivalent. We usually make these types of definitions, and relations because it preserves the information we care about in a set, in this case if all we care about is what the remainder of an integer is when divided by 5.
This is a quite algebraic example, as $\mathbb{Z}/5\mathbb{Z}$ is a field, but different examples as available all throughout mathematics.
A topological example could be if you start with the line segment $[0,1]$ and make the equivalence relation, $1$~$0$, then you are stating that $0$ is equivalent to $1$. Since these two points are now the 'same' to us, what we end up with is a circle.
An equivalence class usually has a fixed representative, usually with some nice feature. Like the equivalence class of all integers with remainder 0 after division by 5, has the representative 5, but we could also use the representative -10, or 1405, 5 is just easier to work with.
Another good example of an equivalence class is when subtraction is initially defined over the integers. You say that $(a,b)$~$(c,d)$ iff $a+d=b+c$, and then we can define subtraction as $a-b:=\{(c,d)|a+d=b+c\}=[(a,b)]$ where $[(a,b)]$ is the usual notation for the equivalence class of $(a,b)$. This equivalence relation basically tells us that 5-4=3-2, since 5+2=4+3, which seems very obvious, but when you're defining subtraction from set theoretical axioms, there is no avoiding this step. I could they say that $(5,4)$ is equivalent to $(3,2)$ which is equivalent to $(n,n-1)$ for any $n\in\mathbb{Z}$. We could also say that $(5,4)\in[(n,n-1)]$ and the contrary is true, since both these objects are symmetrically equivalent to each other (since equivalence relations are symmetrical), so $(n,n-1)\in[(5,4)]$.
A partition is just a way of cutting up a set into disjoint subsets. An equivalence relation always defines a partition. If we split a set up into equivalence classes via some equivalence relation, every element is in some equivalence class, at least its own equivalence class (via reflexivity of equivalence relations). This means that an equivalence relation definitely splits ALL of the set. Are these subsets disjoint though? The answer is yes because if $a$~$b$, then $b$~$a$, by symmetry of an equivalence relation, and if $a\in[b]$ and $a\in[c]$, then $a$~$b$ and $c$~$a$ and by transitivity $c$~$b$, so $[c]=[b]$. In other words if $a$ is in equivalence class of two different elements, then those two elements are equivalent.
Now an equivalence relation always defines a partition, and it's true that a partition always defines an equivalence relation. Just set the relation to be $a$~$b$ iff $a$ and $b$ are in the same region of the partition. You'll find that this relation is infact an equivalence relation.
Finally (sorry for the long answer!), a quotient set $X/R$ where $X$ is some set and $R$ is some relation can be thought of as looking at all the representatives of the equivalence classes of $X$ induced by $R$.
Take the topological example mentioned before. $[0,1]/$~, where $[0,1]$ is our set and ~ is our equivalence relation which sets $0$~$1$. We look at $[0,1]$ is terms of equivalence class of ~ which fortunately for us, only involves 0 and 1, so we pick some representative from the class $[0]=[1]$, say $0$, and now we are looking at a circle.
Take our algebraic example as well. $mathbb{Z}/5\mathbb{Z}$, now $\mathbb{Z}$ is just the integers, and our relation is $a$ and $b$ are equivalent if they differ by a multiple of 5. I should really then write $\mathbb{Z}/$~ where ~ is the above equivalence relation. Then look at the equivalence classes (or the partition of $\mathbb{Z}$) and we find the simple equivalence classes $[0],[1],[2],[3],[4]$, and what the quotienting of the set does, is it says: with some sensible choice of representative, our new quotient set is just the set of representatives. So $\mathbb{Z}/$~ $=\{0,1,2,3,4\}$.
I hope some of this helps. If not have a peak at 
http://en.wikipedia.org/wiki/Equivalence_relation
and
http://mathworld.wolfram.com/EquivalenceRelation.html
wikipedia has a billion examples :)
