Identifications, saturation and gluing

Try

Take the interval [0,1] and identify its two endpoints. The map t↦exp(2πit) traces the unit circle and assigns the same value to exactly the identified endpoints. This suggests a circle quotient, but a picture alone does not establish the topology.

A quotient point represents an entire equivalence class. Any proposed function on the quotient must assign one value to that class, so representative independence is the first proof obligation.

Predict

Predict whether every subset of the original space is the inverse image of some quotient subset. Only unions of full equivalence classes have that property.

Predict also whether gluing well-behaved spaces always gives a Hausdorff space. Identifying corresponding nonzero points of two real lines while keeping their origins separate produces two distinct points that cannot have disjoint neighborhoods.

Experiment

Track endpoint identification on [0,1].

Source subset	Saturated?	Reason
(0,1)	Yes	Contains only complete singleton classes
Singleton 0	No	Omits the equivalent endpoint 1
Pair 0,1	Yes	Contains the whole endpoint class
[0,1/2)	No	Includes 0 but excludes 1

Saturation is a set-theoretic condition. A saturated set need not be open, and an open set need not be saturated.

Understand

For a surjection q, the saturation of A⊆X is q⁻¹(q(A)): it contains every point sharing a fiber with some point of A. Thus A is saturated exactly when this set equals A. Saturated subsets correspond bijectively to subsets of the quotient.

Suppose f:X→Z is constant on each fiber. Define h(q(x))=f(x). Fiber constancy proves well-definedness, and surjectivity gives uniqueness. If q is quotient and f is continuous, the identity h∘q=f and the mapping-out criterion make h continuous.

For endpoint identification, the exponential map on [0,1] has exactly the desired fibers and is onto the circle. It therefore induces a continuous bijection from the quotient to the circle. The later compact-to-Hausdorff theorem proves this bijection is a homeomorphism: the interval and hence its quotient are compact, while the circle is Hausdorff. This last implication is a theorem dependency, not something supplied by bijectivity alone.

For a contrasting construction, start with the disjoint union of two copies of R and identify corresponding nonzero points. Leave the two origins as separate classes. An open neighborhood of either origin has preimage containing an interval about zero in its own copy. Two such intervals overlap at some nonzero coordinate; those copies of the coordinate are identified. Hence every pair of neighborhoods of the two origins intersects. The quotient is not Hausdorff even though both original copies were.

Apply

Determine whether the union of an open set with all equivalent representatives stays open in a given example. In the double-origin construction, saturating an interval from one copy adds its nonzero interval to the other copy, which is open there away from zero. In other equivalence relations saturation need not preserve openness.

Repair “Identifying points only changes the number of points.” It also changes which unions of classes are open and can destroy separation properties. Use the full preimage rule rather than a sketch of glued points.

A fresh collapse: check fibers before drawing conclusions

Take X=[0,1] and 0<a<b<1. Collapse the whole closed interval A=[a,b] to one point, leaving every point outside it separate. A useful coordinate flattens the middle interval while keeping both outer intervals in order:

$$F(x)=\begin{cases}x/(2a),&0\le x<a,\\1/2,&a\le x\le b,\\1/2+(x-b)/(2(1-b)),&b<x\le1.\end{cases}$$

The formulas agree at a and b, so F is continuous; it is onto [0,1], strictly increasing on each outside interval, and has exactly the desired fibers. It therefore induces a continuous bijection from X/A to [0,1]. To prove inverse continuity rather than assuming it, note that X/A is compact as a continuous image of compact X. Every closed subset of this quotient is compact; its image under the induced bijection is compact and hence closed in Hausdorff [0,1]. Thus the bijection is closed, and its inverse is continuous. This argument does not first assume the quotient is Hausdorff.

Does the coordinate identify exactly the intended points?

Identify every point of the selected interval as one class; all other points remain singleton classes. Predict whether the endpoints belong to that class before inspecting the candidate coordinate.

Interval endpointsa=1/4, b=3/4a=1/8, b=1/2 Collapsed setClosed interval [a,b]Open interval (a,b) Approach index n2481664Reset collapse comparison

Red endpoint circles are filled: included in the requested collapsed class. Navy square: x_n=a+(b−a)/n=0.375000, always inside the class. Its distance to a is 0.125000. The curve is the same continuous candidate F in both modes.

F(a)=F((a+b)/2)=1/2. These are the intended same class. The exact piecewise formula has precisely the closed-collapse fibers.

The collapsed fiber is closed. The compact closed-relation proof below establishes Hausdorffness.

Repair the proof

The graph samples 129 values of a known piecewise linear function; the explicit formula proves continuity and fibers. Finitely many sequence values cannot prove a limit or a separation axiom. Use x_n−a=(b−a)/n and the full preimage definition in the written argument.

Twenty bounded settings, temporary state, no automatic mastery. Retain the hypothesis audit and revised proof in the existing written practice.

Now collapse only (a,b), leaving both endpoints separate. The same F is still constant on each requested class, so it descends continuously, but it identifies three distinct quotient points: the class (a,b) and the two endpoint singletons. The quotient is not T1: the preimage of its collapsed singleton is (a,b), not closed in X. Its failure is not caused by a poor choice of coordinate. Every continuous map from this quotient into a Hausdorff target must give the same value at those three points. Indeed its pullback is constant on (a,b); sequences inside that interval converge to each endpoint, and uniqueness of limits in a Hausdorff target forces the endpoint values to agree with that constant.

Closed relations give the general compact separation criterion

Let X be compact Hausdorff, R an equivalence relation in X×X, and q:X→Y its quotient map. Then Y is Hausdorff if and only if R is closed. Here R consists of all pairs of equivalent representatives. Closedness of individual classes alone is only the T1 condition, not this stronger product condition.

First assume R is closed. Each equivalence class is closed: for a fixed x, the continuous map y↦(x,y) pulls R back to that class. For any closed C⊆X, its saturation is the projection onto the first coordinate of R intersected with X×C. This intersection is closed in the compact finite product, hence compact. Its projected image is compact, and therefore closed in Hausdorff X. Since q⁻¹(q(C)) is precisely this saturation, the quotient closed-set criterion shows that q(C) is closed. Thus q is a closed map; no quotient Hausdorff assumption was used to establish this.

Two distinct classes K and H are disjoint closed subsets of X. Compact Hausdorff spaces are normal, so choose disjoint open neighborhoods U and V containing them. Use the proof in compactness and continuous maps here; that later lesson is directly accessible and is the named separation dependency. Raw q(U) and q(V) need not be disjoint or open. Instead form the saturated open cores

$$U^*=X\setminus q^{-1}\bigl(q(X\setminus U)\bigr),\qquad V^*=X\setminus q^{-1}\bigl(q(X\setminus V)\bigr).$$

Each subtracted set is the closed saturation of a closed set, so both cores are open. They are saturated because complements of unions of whole classes are also unions of whole classes. Ucontains K: the class K lies wholly in U and cannot meet X outside U. Similarly V contains H. Also U⊆U and V⊆V by reflexivity of the relation. Thus they are disjoint. Saturation and quotient openness make q(U) and q(V) disjoint open neighborhoods of the two quotient points. This proves Hausdorffness.

Conversely, suppose Y is Hausdorff. Its diagonal is closed: any pair of unequal points has disjoint neighborhoods, whose product misses the diagonal. The relation R is the inverse image of that diagonal under the continuous map q×q. Hence R is closed. This direction does not require compactness, while the forward proof used it for closed projections and normal separation.

For the closed interval collapse, R is the union of the diagonal and A×A, both closed. For the open collapse it is not closed: pairs (m,a+(b−a)/n), with m=(a+b)/2 and n≥2, belong to R but converge to (m,a), which does not. This is an infinite sequence proof, not a claim inferred from the finite visual samples.

Turn the construction into a proof portfolio

Retain a first attempt at the new collapse task. In a second pass label five obligations: exact fibers, continuity upstairs, continuity of the descended map, inverse continuity, and separation. Return to the existing quotient, compactness and separation lessons for any failed step. Revise the proof after testing the open-collapse counterexample, and ask a reviewer to challenge one unstated hypothesis. Record an actual review only when someone has supplied it; a saved draft or opened solution is not independent mathematical evidence.

The MIT 18.901 Fall 2004 calendar includes quotient and separation constructions along with later extension, compactification and dimension topics. The assignment guidance requires revised proof writing. The present collapse and criterion are original practice joining those capabilities, not a reproduction of an inaccessible textbook exercise. Existing Tietze and Stone–Čech proofs are reused elsewhere in the assessment; the full finite-dimensional embedding theorem remains a separate dependency.

Continue

The next lesson gives precise names to point-separation properties and explains the double-origin obstruction. Optional depth: construct a map from the endpoint quotient to any space by specifying a continuous interval map whose endpoint values agree. The quotient universal property then supplies continuity without inspecting quotient neighborhoods individually.