Summary of "Введение в дискретную математику и топологию. Лекция 8. Артамкин И. В. 05.03.2026"

Summary — main ideas, concepts and results

Context and administrative notes

Lecturer: Artamkin I. V.
Course logistics:
- Grades for task 4 are uploaded to the LMS (some students see 0/10 — check for upload or technical errors).
- The last individual assignment (number A) will open later and is extended until next Saturday.
- A final (optional) lecture on an extra topic (recursive sequences and generating functions) may be given separately (likely Mon 16th or Wed 11th, to be confirmed).
- The make-up colloquium for the module is scheduled for Tuesday the 24th; students will be assigned to one of three time blocks.
Administrative names mentioned:
- Solomon (decision about the last individual assignment)
- Osif (Osip) Vladimirovich (another lecturer whose attendees may request time adjustments)
- Academic department

Topological preliminaries and notation

Reminder of the Jordan curve theorem (plane version):

A simple closed curve C (continuous injective image of S1 in R2) splits R2 into exactly two connected open components whose common boundary is C.
Basic topological definitions recalled:
- interior: largest open set contained in X
- closure: smallest closed set containing X
- boundary: closure(X) \ interior(X)
Emphasis: always check which notion of boundary is intended in statements.

Applications and examples

1) The “two people / two carts” configuration-space argument

Problem sketch: two people tied by a fixed-length rope walk along two given paths from A to B and B to A (or cars of radius > half-rope length) — show they cannot be separated (i.e., a collision/overlap-free motion exists).
Modeling idea:
- Use configuration space X1 × X2 (a rectangle/square when each path is parametrized by an interval).
- The simultaneous positions of the two moving objects trace a continuous curve in the square from (A,B) to (B,A).
- Using a Jordan-like separation argument one shows an avoidance is impossible (the curves must intersect or cross the forbidden diagonal region).
Sketch of the topological lemma used:
- Label four points A, B, C, D on a Jordan curve in parameter order.
- Consider two arcs/paths inside one region connecting A–C and B–D.
- Construct a new closed curve by gluing an arc of the original circle with one interior path.
- Apply the Jordan theorem and a locally constant function (taking different constants on the two sides) to reach a contradiction unless the interior paths intersect.

2) Brouwer fixed-point theorem — statement and two proof sketches

Theorem (Brouwer):

Any continuous map f : D^n → D^n from the closed n-ball to itself has a fixed point (f(x) = x for some x).
1D case: immediate via the intermediate value theorem using g(x) = f(x) − x.
Two approaches sketched for 2D and higher:

A. Combinatorial / Sperner-based proof (2D specialization)

Idea (by contradiction): assume f has no fixed point. For each x, consider the ray from f(x) through x; that ray meets the boundary circle at a point — define g(x) as that boundary intersection. This gives a continuous map g : D^2 → S1 with g|_{S1} = id.
Partition the boundary circle S1 into three arcs V1, V2, V3 and set Wi = g^{-1}(Vi). The Wi cover the disk. Triangulate the disk (via a homeomorphism to a triangle) and label vertices 1/2/3 according to which Wi they lie in, with boundary labeling matching the three arcs.
Apply Sperner’s Lemma:
- Sperner’s lemma (2D): in any triangulation of a triangle whose vertices are labeled 1,2,3 and whose boundary labeling respects the three sides (each side only uses two labels), there is at least one small triangle with labels 1,2,3.
Refining triangulations gives a sequence of 1-2-3 triangles; by compactness a limit point lies in the triple intersection of closures, contradicting the assumption. Hence f must have a fixed point.

Sketch of Sperner’s lemma proof (parity / “doors” argument):

Treat edges between vertices labeled 1–2 as “doors”.
The selected boundary side between labels 1 and 2 has an odd number of such doors.
Following doors through adjacent triangles: passthrough triangles have two doors, while a 1-2-3 triangle yields a single-door dead end.
Parity/finite path argument implies existence of at least one 1-2-3 triangle.

B. Topological / mapping-space / algebraic-topology-minded proof (conceptual)

Introduce pi0(X): the set of path-connected components of X; a continuous map f : X → Y induces f_* : pi0(X) → pi0(Y).
Mapping spaces: for fixed Z, composition by f gives f_* : C(Z, X) → C(Z, Y) (C denotes continuous maps Z → ·). For compact Z and metric spaces, this composition is continuous (technical details deferred).
For Z = S1, components of C(S1, S1) correspond to homotopy classes / integers: maps S1 → S1 are classified by degree (winding number). Thus C(S1, S1) has countably many connected components indexed by Z.
Applying this to the Brouwer setup: the constructed g : D^2 → S1 restricts to identity on S1, so composition of induced maps on mapping-space components would imply an impossible identity between multiple components and a single component (equivalently, it would produce a continuous retraction D^2 → S1). The mismatch in component structure (degree/winding-number obstruction) gives a contradiction. The lecturer promised to formalize the degree argument in the next lecture.

Key definitions, tools and invariants used

Jordan curve theorem (plane).
Interior, closure, boundary (with caution about intended meaning).
Configuration space modeling (product of path-parameter spaces).
Locally constant functions on disconnected sets and continuity contradictions on connected curves.
Triangulation and mesh-refinement (simplicial decomposition) argument.
Sperner’s lemma (combinatorial topological lemma).
pi0 (set of path-connected components) as a functorial invariant.
Mapping spaces C(Z, X) and induced composition maps.
Degree / winding number of circle maps S1 → S1, classifying path-components of C(S1, S1).
Non-existence of a continuous retraction from D^n onto S^{n-1} (used implicitly).

Methodologies / procedural outlines (detailed steps)

A) Configuration-space intersection (two-people / two-carts lemma)

Parametrize each path by an interval; denote positions x1(t), x2(t).
Consider the product parameter space (square) where a joint motion traces a continuous curve from (A,B) to (B,A).
Show any continuous curve connecting (A,B) and (B,A) must intersect the “forbidden” diagonal region corresponding to collisions.
If needed, construct a closed curve by gluing an arc of the boundary with an interior path, apply Jordan’s theorem, define a locally constant function that takes different constants on the two complementary regions, and use continuity on the connected parametrized curve to force a point of intersection.

B) Sperner-based combinatorial proof of Brouwer’s fixed-point theorem (2D)

Assume f : D^2 → D^2 has no fixed point.
Define g(x) as intersection of the ray from f(x) through x with S1; then g is continuous and g|_{S1} = id.
Partition S1 into three arcs V1, V2, V3. Let Wi = g^{-1}(Vi); these cover the disk.
Map the disk homeomorphically to a triangle and triangulate; label each vertex by i ∈ {1,2,3} according to Wi, with boundary vertices labeled so each side uses only two labels.
Apply Sperner’s lemma to deduce existence of a small triangle labeled 1,2,3 in each triangulation.
Refine triangulations; take a convergent subsequence of the centers of these 1-2-3 triangles (compactness) to obtain a point in the triple intersection of closures — contradiction.
Conclude f has a fixed point.

C) Sperner’s lemma proof (parity / “doors”)

Label triangulation vertices by 1/2/3 with boundary conditions.
Count edges on a selected boundary side that connect 1–2: an odd number results.
Interpret edges as doors; follow paths through triangles: passthrough triangles have two doors; only a 1-2-3 triangle yields a single-door dead end.
Parity argument implies existence of a 1-2-3 triangle.

D) Topological / mapping-space approach (conceptual steps)

For continuous f : D^n → D^n, if no fixed point exists, construct continuous g : D^n → S^{n-1} sending x to the boundary intersection of the ray from f(x) through x.
Then g|_{S^{n-1}} = id (g is a retraction).
Consider mapping spaces C(S^{n-1}, S^{n-1}); their path-components reflect invariants (for n = 2, components correspond to integer degree).
Existence of a retraction would induce impossible behavior on components (e.g., collapsing nontrivial components to a single one and back), contradicting invariants like degree. Conclude a fixed point must exist.

Speakers / sources featured

Lecturer: Артамкин И. В. (Artamkin I. V.)
Theorems and classical sources referenced:
- Jordan Curve Theorem (Camille Jordan)
- Brouwer Fixed-Point Theorem (L. E. J. Brouwer)
- Sperner’s Lemma (Eduard Sperner)
Other administrative mentions:
- Solomon
- Osif (Osip) Vladimirovich
- Academic department

Notes / follow-up

The lecturer indicated the mapping-space / degree arguments (classifying components of C(S1, S1) by integer degree) will be formalized in the next lecture.
A seminar exercise related to the ray-projection lemma was promised.