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Use the quotient rule
Chapters 6–8. Distinguish continuity, openness and descended maps.
Full chapter index · Practice and worked checks · Topology subject root
Join the endpoints of an interval and the resulting space behaves like a circle. But an open set near just one endpoint need not become an open set after the join.
The missing step is saturation: when points are identified, a set representing complete quotient points must include every representative of each one. This guide tests the seam with explicit sets and then uses the quotient rule to check functions.
Begin with The Gold Standard of Topology and How Mathematics Works: Topology. We work with the usual topology on X = [0, 1] throughout.
Chapter index
Define the quotient · Chapters 1–2
1. State exactly which points are joined
2. Use the full inverse-image test
Inspect the seam · Chapters 3–5
3. Define saturation before testing an image
4. Follow one open half-neighbourhood to the seam
5. Supply both sides of the quotient neighbourhood
Use the quotient rule · Chapters 6–8
6. Distinguish a quotient map from an open map
7. Check that the quotient really is a circle
8. Decide whether a function descends through the join
Check and practise · Chapters 9–10
9. Change the identification and recompute saturation
10. Practise with a new set and a new function
Read and continue · Chapters 11–12
11. Questions readers ask
12. Sources and useful reading routes
Define an equivalence relation on X that identifies 0 with 1 and makes every other point equivalent only to itself.
The quotient set Q consists of these equivalence classes. The endpoint class is {0, 1}; every interior class is a singleton {t}.
Let q map each t to its class [t]. It is surjective: every quotient point has at least one representative.
Drawing a circle suggests the construction, but the quotient topology still needs a definition. The set of classes alone does not specify which subsets are open.
The distinction resembles choosing a topology for an infinite product: the underlying set and the rule for open sets are separate pieces of information.
A subset U of Q is open in the quotient topology precisely when q⁻¹(U) is open relative to X.
Relative openness matters. For example, [0, 0.1) is open in X because it equals X intersected with the open real interval (−0.1, 0.1).
The rule uses all representatives of the quotient points in U. If U contains the endpoint class, its inverse image contains both 0 and 1.
Continuity of q follows directly: the inverse image of every open quotient set is open in X. A stronger equivalence defines the quotient topology, because openness of the inverse image also guarantees openness of U.
For A contained in X, its saturation is Sat(A) = q⁻¹(q(A)). It adds every point equivalent to a point already in A.
If A contains 0 but not 1, saturation adds 1. If it contains 1 but not 0, saturation adds 0. Interior points add no different representatives in this example.
A is saturated when it is a union of complete equivalence classes, or equivalently Sat(A) = A.
To decide whether q(A) is open, test Sat(A), not merely A. The quotient rule says q(A) is open if and only if Sat(A) is open in X.
This extra preimage step is where a tempting open-image shortcut can fail.
CHAPTER 4 OF 12 · Inspect the seam
4. Follow one open half-neighbourhood to the seam
Take A = [0, 0.1). It is open in X.
Its image contains the joined endpoint and a short arc on one side. Its full inverse image is [0, 0.1) ∪ {1}.
This saturation is not open in X. Every relative open neighbourhood of 1 includes points just below 1, and those points are absent.
Therefore q(A) is not open in Q, even though A was open in X.
The calculation does not contradict continuity. Continuity concerns inverse images of open target sets. It does not require images of arbitrary open source sets to be open. A map with that latter property is called an open map.
CHAPTER 5 OF 12 · Inspect the seam
5. Supply both sides of the quotient neighbourhood
Now use B = [0, 0.1) ∪ (0.9, 1]. Both pieces are open relative to X.
B contains both representatives of the endpoint class, so it is saturated. Consequently q⁻¹(q(B)) = B is open, and q(B) is open in Q.
| Source set | Saturation | Open in X? | Image open in Q? |
|---|---|---|---|
| [0, 0.1) | [0, 0.1) ∪ {1} | Source yes; saturation no | No |
| [0, 0.1) ∪ (0.9, 1] | Same set | Yes | Yes |
| (0.4, 0.6) | Same set | Yes | Yes |
| {0, 1} | Same set | No | No |
The two seam pieces need not have equal lengths. They need to provide relative neighbourhoods of both endpoint representatives.
A picture of only one representative can hide half the required neighbourhood.
CHAPTER 6 OF 12 · Use the quotient rule
6. Distinguish a quotient map from an open map
Every open saturated source set has an open image under a quotient map: its saturation is itself, so the defining test applies.
That does not imply every open source set has an open image. Our set A supplies a counterexample for q.
A continuous surjective open map is a quotient map. To verify the reverse openness implication, suppose q⁻¹(U) is open; surjectivity gives q(q⁻¹(U)) = U, which is open because q is an open map.
The converse fails: the interval endpoint map is quotient but not open.
These are different hypotheses, not different names for the same condition. Keeping them separate prevents a false step in a gluing argument.
CHAPTER 7 OF 12 · Use the quotient rule
7. Check that the quotient really is a circle
Define F(t) = (cos 2πt, sin 2πt). It maps X continuously onto the unit circle S¹ and identifies precisely the endpoint pair.
Because F is constant on each equivalence class, it defines a bijection g from Q to S¹ by g([t]) = F(t).
The quotient property makes g continuous: for an open set O in S¹, q⁻¹(g⁻¹(O)) = F⁻¹(O) is open.
Q is compact as the continuous image q(X) of a compact interval. The circle is Hausdorff. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, so g has a continuous inverse.
This checks the circle picture through stated hypotheses. The continuous-bijection counterexample explains why bijectivity and continuity alone would not suffice in general.
CHAPTER 8 OF 12 · Use the quotient rule
8. Decide whether a function descends through the join
A function h on X defines a function on Q only if h(0) = h(1), because the quotient treats those endpoints as one point.
When h is continuous and has that endpoint agreement, the induced function h̄([t]) = h(t) is well defined and continuous. Its composition with q is h; the quotient test transfers continuity.
| Function on [0,1] | Endpoint agreement? | Continuous function on Q induced? |
|---|---|---|
| h(t) = t | No: 0 differs from 1 | No well-defined induced function |
| h(t) = t(1 − t) | Yes: both zero | Yes |
| h(t) = cos 2πt | Yes: both one | Yes |
Agreement alone ensures well-definedness, not continuity of an arbitrary h. The continuity assumption has its own job.
This is a useful way to test whether a formula respects the identification before trying to prove anything about the resulting space.
CHAPTER 9 OF 12 · Check and practise
9. Change the identification and recompute saturation
Identify 0, 1/2 and 1 as one class, leaving other points separate. This is a different quotient.
For A = [0, 0.1), saturation becomes A ∪ {1/2, 1}. It is not open: there are isolated included representatives at 1/2 and 1 without suitable surrounding source points.
An open saturated set containing that class must include a relative neighbourhood around every representative, including both sides of 1/2.
The earlier endpoint calculation cannot be copied unchanged. The equivalence classes have changed, so the saturation calculation changes.
Do not infer the quotient’s full global shape from this one test. The exercise establishes a particular open-set requirement, which can then support a fuller analysis.
CHAPTER 10 OF 12 · Check and practise
10. Practise with a new set and a new function
Return to the original endpoint identification. Let C = [0, 0.2) ∪ (0.95, 1].
It is open relative to X and saturated. Thus q(C) is open even though the two seam pieces have different lengths. Symmetry was never a requirement.
Next test D = [0, 0.2) ∪ [0.95, 1]. It is saturated, but not open because 0.95 has no small relative neighbourhood contained in D. Therefore q(D) is not open.
Finally test h(t) = sin πt. It is continuous and has h(0) = h(1) = 0, so it induces a continuous function on Q.
Change it to h(t) = t + sin πt. The endpoint values become zero and one, so well-definedness fails before continuity can be considered.
Is saturation the same as openness?
No. The set {0,1} is saturated but not open in X. The half-neighbourhood [0,0.1) is open but not saturated.
Does a quotient map have to be one-to-one?
No. It is surjective, and may identify several source points. The entire purpose here is to identify two endpoints.
Can I test only one representative of a seam point?
Not for the inverse-image openness rule. The preimage contains all representatives of the quotient point.
Does continuous gluing preserve every property?
No universal conclusion follows. Each property and theorem requires its own hypotheses. The quotient topology gives a precise continuity framework, not an automatic preservation rule for all spaces.
Quotient definitions and gluing
Allen Hatcher’s introductory point-set topology notes, Quotient Spaces develop the inverse-image definition and the need to include all representatives near a seam. The explicit set tests and function table above work through that framework on the stated interval.
Follow nearby distinctions
Compare a continuous bijection with a discontinuous inverse, connectedness without path connectedness and coordinatewise convergence in infinite products. Each illustrates why a definition’s missing condition can change the conclusion.
Return to the learning map
Use Topology, the topology mechanism guide, The Gold Standard of Education and How X Works. Identify the classes, lift the complete set, then test the required property.
Sources checked 22 October 2026. Worked examples are educational models with the assumptions stated above.