SYSTEMATIC MATHEMATICS

Algebraic Topology

How can a loop become a group element, how can boundaries become matrices, and how can holes be compared across maps, products, manifolds, and noisy finite data? This course answers those questions in forty-four visible bilingual chapters across nine content-sized units of lengths 5, 5, 6, 5, 5, 5, 5, 4, and 4. It begins with homotopy, fundamental groups, and covering spaces; constructs simplicial, singular, relative, and cellular homology; develops exact sequences, excision, Mayer–Vietoris, cohomology, cup products, coefficient theorems, manifold duality, fixed-point evidence, persistent homology, higher homotopy, fibrations, and spectral-sequence reading; and ends with an independently reconstructable torus dossier. Every basepoint, orientation, coefficient group, boundary sign, induced map, quotient, and model assumption is named. The course does not present one invariant as a complete classification, confuse homotopy equivalence with homeomorphism, or claim that a barcode proves a physical feature without a validated filtration and stability argument.

Before this course: Completed Topology, Abstract Algebra, Proof, Logic & Set Theory, and Linear Algebra. Students should already know continuous maps, quotient spaces, compactness, connectedness, path homotopy, groups, normal subgroups, quotient groups, free abelian groups, bases, matrices, kernels, images, and proof by universal properties. Differential Geometry & Manifolds is useful for the final duality bridge but is not required. No computational-topology package, point-cloud model, category theory course, or spectral-sequence fluency is assumed.

COURSE FACTSLevel, chapters, units, prerequisite, and outcome
Chapter 1

Topological invariants answer declared questions, not every question at once

Objective: Why does equality of one invariant not prove two spaces homeomorphic?

A homeomorphism preserves every topological property, but proving two spaces are not homeomorphic directly can be difficult. An invariant assigns algebraic data that homeomorphisms must preserve, so one unequal value can disprove equivalence. First name the spaces, maps, basepoints, coefficient group, and equality or equivalence relation. The chapter’s exact claim is: If I(X) and I(Y) are not isomorphic, then X and Y are not homeomorphic; the converse need not hold.

The object being computed is: A topological invariant I assigns an object I(X) to each space so that every homeomorphism f:X→Y induces an isomorphism I(f):I(X)→I(Y). A complete invariant also has the converse property; most invariants in this course are not complete. Algebraic topology replaces a geometric deformation problem by algebra only after proving that the algebra is invariant under the allowed deformation.

Use the exact problem “Use the number of connected components to compare X=[0,1]∪[2,3] with Y=[0,1]. Can they be homeomorphic, and what remains unproved?” to rebuild the construction rather than memorize a named theorem. Check every boundary map or induced map, verify the result in a second way, and test this limitation: The interval [0,1] and a point are homotopy equivalent but not homeomorphic. A homotopy invariant cannot distinguish them as homeomorphism types.

If I(X) and I(Y) are not isomorphic, then X and Y are not homeomorphic; the converse need not hold.

Assume a homeomorphism f:X→Y existed; invariance would provide an isomorphism I(f).

Nonisomorphic I(X) and I(Y) make such an isomorphism impossible.

Therefore the assumed homeomorphism cannot exist; equal invariants merely leave the question undecided.

Use the number of connected components to compare X=[0,1]∪[2,3] with Y=[0,1]. Can they be homeomorphic, and what remains unproved?

  1. X has two connected components because its two intervals are separated; Y has one.
  2. A homeomorphism maps connected components bijectively, so the component counts must agree.
  3. The unequal counts disprove homeomorphism, but the invariant says nothing by itself about other pairs having the same count.

Result: They are not homeomorphic because 2≠1 components; component count is an obstruction, not a complete classification.