What is an intuitive explanation of a quotient space?

Quotienting is smooshing stuff together. In ordinary multiplication you count groups of equivalent things. For example, 5 groups of 3 students each. The students are not all alike in every way, but they're alike for our purposes. Maybe everyone in the group shares a classroom chore. When you quotient you then focus on the circles in the lower picture rather than the individual roses. There are three circles. If it were the classroom above, after quotienting I would be talking about 5 groups rather than 15 students. So we obtain quotient spaces by equivalence-classing: identifying some criterion ("all students that are part of Group Rhino") and then smushing them all together for some purpose. Here are a few examples: there are 88 keys on a standard piano. However in some ways A440 is equivalent to A880, and to A220: The pitches are obviously not the same, but modulo getting rid of the different registers they are. The equivalence relation would be to make all numbers that are some power of two multiplied from each other an equivalence class. That quotient operation reduces the 88 keys of the piano to C Dâ™­ D Eâ™­ E F Gâ™­ G Aâ™­ A Bâ™­ B = the chromatic scale. Starting from the complex plane, you could make a half-plane by equivalence-classing either every number with its opposite, or every number with its conjugate (interchanging √−1⟷−√−1−1⟷−−1\sqrt{-1} \longleftrightarrow -\sqrt{-1}). Anytime someone speaks in generalities, such as "the poor are smart" or "the poor lack conscientiousness", they are talking about an equivalence class of people rather than individuals. from http://isomorphismes.tumblr.com/tagged/lie+groups: (pics by John Starett) you might want to quotient a space of curves (for example solutions to some system of equations that's important to you) by a natural symmetry. In the pictures above ∃ a symmetry to exploit which can simplify solving some ODE's. Wikipedia's article on http://en.wikipedia.org/wiki/Orbifolds takes the idea of using a symmetry to shrink something down even further, noting: The underlying space locally looks like the https://en.wikipedia.org/wiki/Quotient_space of a https://en.wikipedia.org/wiki/Euclidean_space under the https://en.wikipedia.org/wiki/Linear_map https://en.wikipedia.org/wiki/Group_action of a https://en.wikipedia.org/wiki/Finite_group. ... https://en.wikipedia.org/wiki/Automorphic_form ... geometry of https://en.wikipedia.org/wiki/3-manifold ...https://en.wikipedia.org/wiki/CAT%28k%29_space i.e., different ways of quotienting lead to interesting mathematical structures. I believe also the concept of foliations of a manifold comes from quotienting the manifold. gave a nice answer in describing the quotient operation in the abstract. In http://isomorphismes.tumblr.com/tagged/combinatorics, you can start off with the number of permutations of a set n!(n−k)!n!(n−k)!n! \over (n-k)!—for example the number of possible lottery tickets, http://www.evolutionfaq.com/articles/probability-life, the number of possible DNA strings, the number of ways gas molecules could be arranged in a balloon—but then maybe you realise order does not matter. So you quotient off or modulo the number of possible rearrangements /k!/k!/k! and that's the relationship between combinations and permutations. You've quotiented by a symmetric group like in Starrett's pictures above. Except instead of being a planar situation or manifold situation, this is more of a "pure numerical" situation or rougher discrete structure. Say you are and want to assign probabilities like 37%, 22%, and 42%, rather than giving 100 separate events each corresponding to one face of the die. So you make an equivalence-class of 37 faces together and assign the 37% outcome to that equivalence-class, rather than assigning the same outcome to the 37 faces individually. (Veering off track: Dungeons & Dragons players know that a 100-sided die can be simulated with two 10-sided dice—this is the Cartesian product operation—so potentially there's something to be said about Cartesian-producting and then quotienting. Also NB that we could count (combinatorics again) the number of ways to equivalence-class 37 faces out of 100. Using maths on itself. Anyway....) Philosophically, Willard van Orman Quine http://isomorphismes.tumblr.com/post/2747507384/quine the integer 5 to be "the equivalence class of all sets with cardinality 5". You could take the real line RR\mathbb{R} and modulo by one. Then you have the equivalence class isomorphic to [0,1)∈R[0,1)∈R [0,1) \in \mathbb{R} . Members of [0,1)[0,1)[0,1) can be added to members of NN\mathbb{N} to recover RR\mathbb{R}. So a computer doesn't need to represent 25.293487298372 separately from .293487298372 — it can do its post-decimal arithmetic separately from its whole-number arithmetic. You could take the real line RR\mathbb{R} and modulo by sign. You then have R+∪{0}R+∪{0}\mathbb{R}^+ \cup \{ 0 \}. You could take the real line RR\mathbb{R} and modulo by sign, then take the equivalence class by x↦⌊x⌋x↦⌊x⌋x \mapsto \lfloor x \rfloor . This quotient is the natural numbers (including zero). Going outside my expertise, there may be a way to start with the real line RR\mathbb{R}, or a https://en.wikipedia.org/wiki/Gunk_%28mereology%29  line, or a non-archimedean field, and quotient by some kind of infinitesimal quantity, and get back to a "less smooth" level of density (something more spiky or separated). The rational numbers QQ\mathbb{Q} are "overloaded" by just taking ratios of integers, i.e. Z×Zonto−−→QZ×Z→ontoQ\mathbb{Z} \times \mathbb{Z} \xrightarrow{\rm onto} \mathbb{Q} ratios (pairs) are surjective, but don't injectively map onto QQ\mathbb{Q}. The overlap would be of equivalent (like "enharmonic") ratios like 24=1632=193824=1632=1938{2 \over 4} = {16 \over 32} = {19 \over 38}. Which is why you will always see, when someone writes about a ratio pqpqp \over q of integers, they have to put the qualifier that this is "reduced form". If you think about the family of lines basing out of (0,5)(0,5)(0,5) namely y=mx+5y=mx+5y=mx+5, and you think about the family of lines of equal slope like y=3.2x+by=3.2x+by=3.2x+b, and you think about then the space of all such possible lines in the plane parameterised by (m,b)∈R×R(m,b)∈R×R(m,b) \in \mathbb{R} \times \mathbb{R}--then you could overparameterise that by defining lines basing out of any (x,y)(x,y)(x,y) point along with a slope m∈[0,∞)m∈[0,∞)m \in [0,\infty)----and all of these families of lines are related by quotient operations. What's the difference between the two-lightswitch group pic by Nathan Carter ↑=↓−1=(↑−1)−1↑=↓−1=(↑−1)−1 \uparrow \quad = \quad \downarrow^{-1} \quad = \quad {\left( \uparrow  ^{-1}\right)}^{-1}  â†=→−1=(←−1)−1←=→−1=(←−1)−1    \leftarrow \quad = \quad \rightarrow^{-1} \quad = \quad {\left( \leftarrow  ^{-1}\right)}^{-1}     and the free group on two generators  aba−1b−1aaaba∈F2aba−1b−1aaaba∈F2   aba^{-1}b^{-1}aaaba \in F_2   aba−1b−1baab=aba−1−aab=ab−−ab∈F2aba−1b−1baab=aba−1−aab=ab−−ab∈F2    aba^{-1}b^{-1}baab = aba^{-1} \!\! - \!  aab = ab \!\! - \! -  ab \in F_2      ? Well in the free group bbbbbbbbbbbb is considered going (let's say north) four steps. But with a lightswitch if you keep hitting ↑↑↑↑↑↑ you will not turn the light on brighter; it's already in the "on" position. In mathematical terms ↑ is idempotent, i.e. ↑↑ = ↑ so ↑↑↑↑↑↑↑↑…↑ = ↑. Squishing all of the bbbbb=↑↑↑↑↑bbbbb=↑↑↑↑↑bbbbb=\uparrow \uparrow \uparrow \uparrow \uparrow with idempotence is a quotient operation. The symmetric group of order 3 operates on ordered triples like (B,A,C)(B,A,C)(B,A,C) and (C,A,B)(C,A,B)(C,A,B) or, equivalently, on triangles—if the triangles are thought of in a certain way. top right. image again by Nathan Carter It's easiest for me to describe precisely what "http://isomorphismes.tumblr.com/post/12848094184/space" the group is operating on using quotients. Start with the set of all triangles in the flat infinite plane—scalene, isosceles, obtuse, equiangular—and quotient away all the locations (eg, quotient away the barycentres). Now these are just triangles not located anywhere. Next quotient away all the (rotational) orientations of the triangles—picking "12 o'clock / north" to be the "top" i.e. whatever angle gets the name "one" or "a" or some such [[[implicit in this I guess we're quotienting away all the names of the angles and sides, fixing them as ABC or whatever]]] is at that top or north spot. Last, quotient away all the inner angles: now it doesn't matter whether it was isosceles or scalene or whatever. But we are still left with side names and angle names, with one being top, one being left, one being right. (It has to be roughly this way by all the quotienting done before.) Drop the side names and now these "abstracted" triangles or equivalence-classes of triangles are what's isomorphic to ordered triples \sym(1,2,3)\sym(1,2,3)\sym (1,2,3). https://en.wikipedia.org/wiki/Pullback we could do operations such as flipping on the original Euclidean plane and these would correspond to group operations in the heavily quotiented space. Or https://en.wikipedia.org/wiki/Pushforward from the original infinite flat plane to the quotiented space and doing interchange σ1,2σ1,2\sigma_{1,2} or shift σ3,1,2σ3,1,2\sigma_{3,1,2} operations on the quotiented thing would equally induce something more complicated, but equivalent to a group operation, on the Euclidean plane itself. Back to music. We could take the 88 keys and drop all of the black ones (project to the white keys). We've now chosen the key of C. Quotient away the octaves and stow this aside for a moment. ... Back to the 88 keys again, we could have chosen any diatonic scale out of the chromatic basis. For example the key of D has D E F F♯ G A B C C♯. Now let's do the same exercise as the white keys except do it for every possible key (even enharmonics? well, I don't really care if you do or don't). Now take the equivalence-class of all of these "white-key" substitutes or diatonic scales, you now have do re mi fa sol la ti. The even numbers are the equivalence class of integers which modulo-2 to zero.

Constructing quotient spaces is a two step procedure: 1) partition the set of points. Each element of the partition becomes a point if the new space. Think of it as a coarser view of the original space (like zooming out with Google maps which collapses streets etc to a single city, cities to prefectures, prefectures to countries,etc, but also allowing zooming out at different speeds depending on region, folding,...) 2) deal with the structure. Let's use topological spaces as an example. How to decide what topology to use. The categorical point of view helps here. It says that the quotient space Q induced by a partition/equivalence relation on X is given by a map X -> Q from your space X to Q which is universal (=best in some sense) among all maps which collaps all equivalence classes, i.e. which map all points from an equivalence class to a single point. More precisely: if you have any map X->A with this property then you can factor this map through X->Q which means X->A = X->Q->A  for some map Q->A. So what topology to choose? It turns out that using all subsets of Q whose preimage is open in X is just right. If you take more open sets then X->Q is not continuous if you choose less then the universal property is no longer satisfied. If you restrict the spaces to some special subclass or change the allowable maps you get different answers to the question of what the quotient space is. But the universal property is always the same. So understanding quotient spaces means (i) understanding the universal property and (ii) figuring out what this universal property implies in a given situation.