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topological invariance of dimension

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

Contents

Idea

A fundamental theorem of topology: for n 1,n 2n_1, n_2 \in \mathbb{N} two different natural numbers, n 1n 2n_1 \neq n_2, then the Cartesian spaces n 1\mathbb{R}^{n_1} and n 2\mathbb{R}^{n_2} of dimension n 1n_1 and n 2n_2, respectively, are not homeomorphic (while for n 1=n 2n_1 = n_2 then of course they are). Hence dimension of topological manifolds is a topological invariant.

This seems intuitively obvious, but the formal proof (due to Brouwer around 1910) is comparatively hard, and the problem was an important open problem in the 19th century. To appreciate that the problem is harder than it may superficially seem it serves to notice that for every nn \in \mathbb{N} there are surjective continuous maps 1 n\mathbb{R}^1 \to \mathbb{R}^n (the Peano curves), in addition to the obvious injective continuous maps 1 n\mathbb{R}^1 \to \mathbb{R}^n for n1n \geq 1 (or indeed m n\mathbb{R}^m \to \mathbb{R}^n for mnm \leq n), because of which there are also bijective (but discontinuous) maps m n\mathbb{R}^m \to \mathbb{R}^n for m,n1m, n \geq 1 (by the Schroeder–Bernstein theorem).

Statement

Theorem

(topological invariance of dimension)

For n 1,n 2n_1, n_2 \in \mathbb{N} then the Euclidean space n 1\mathbb{R}^{n_1} and n 2\mathbb{R}^{n_2} (regarded as topological spaces with their metric topology) are homeomorphic if and only if n 1=n 2n_1 = n_2.

More generally:

Theorem

(Brouwer invariance of domain theorem)

For U nU \subset \mathbb{R}^n an open subset of a Cartesian space, for any nn \in \mathbb{N}, and for f:U nf \colon U \to \mathbb{R}^n a continuous injective map, then its image f(U)f(U) is also an open subset.

In other words, continuous injections between Cartesian spaces of the same dimension are open maps.

This implies theorem 1:

Lemma

For n 1,n 2n_1, n_2 \in \mathbb{N}, there is a continuous injection from n 1\mathbb{R}^{n_1} to n 2\mathbb{R}^{n_2} if and only if n 1n 2n_1 \leq n_2.

Proof

If n 1n 2n_1 \leq n_2, use ι n 1,n 2:(x 1,,x n 1)(x 1,,x n 1,0)\iota_{n_1,n_2}\colon (x_1, \ldots, x_{n_1}) \mapsto (x_1, \ldots, x_{n_1}, \vec{0}), where 0\vec{0} consists of n 2n 1n_2 - n_1 copies of 00. If n 1>n 2n_1 \gt n_2, then supposing a continuous injection f: n 1 n 2f\colon \mathbb{R}^{n_1} \to \mathbb{R}^{n_2}, compose ff with ι n 2,n 1\iota_{n_2,n_1} to get a map from n 1\mathbb{R}^{n_1} to itself, also a continuous injection. By invariance of domain (theorem 2), the image of this map is open in n 1\mathbb{R}^{n_1}, yet contained in the range of ι n 2,n 1\iota_{n_2,n_1}, and the only open subset of that range is empty, a contradiction since n 1\mathbb{R}^{n_1} is not empty.

Proof

of theorem 1 from theorem 2

A homeomorphism is a continuous injection both ways, so n 1n 2n_1 \leq n_2 and n 2n 1n_2 \leq n_1 by lemma 1.

This only-if half of this argument immediately generalizes to any inhabited open subsets of n 1\mathbb{R}^{n_1} and n 2\mathbb{R}^{n_2}.

Proofs

We discuss various proofs of the topological invariance of dimension (theorem 1).

Via ordinary cohomology

Proof

of theorem 1

Consider relative ordinary cohomology H H_\bullet with coefficients in, says, the integers \mathbb{Z}. Recall that for AXA \hookrightarrow X an inclusion of CW-complexes, then there is a long exact sequence of the form

H n1(X)H n1(A)H n(X,A)H n(X) \cdots \to H^{n-1}(X) \longrightarrow H^{n-1}(A) \longrightarrow H^n(X,A) \longrightarrow H^n(X) \to \cdots

Applied to the point inclusion {0} n\{0\} \hookrightarrow \mathbb{R}^n and using that

  1. H k( n)={ |ifk=0 0 |otherwiseH^k(\mathbb{R}^n) = \left\{ \array{ \mathbb{Z} & \vert\, \text{if}\, k = 0 \\ 0 & \vert\, \text{otherwise}} \right.

    because n\mathbb{R}^n is homotopy equivalent to the point;

  2. H k( n\{0})={ |ifk{0,n1} 0 |otherwiseH^k(\mathbb{R}^n \backslash \{0\}) = \left\{ \array{ \mathbb{Z} &\vert\, \text{if} \, k \in \{0,n-1\} \\ 0 & |\, \text{otherwise} } \right.

    because n\{0}\mathbb{R}^n \backslash \{0\} is homotopy equvaent to th (n-1)-sphere S n1S^{n-1}

this gives for n2n \geq 2 that H k( n, n{0})={ |ifk{0,n} 0 |otherwiseH^k(\mathbb{R}^n , \mathbb{R}^n - \{0\}) = \left\{ \array{ \mathbb{Z} &\vert\, \text{if} \, k \in \{0,n\} \\ 0 & \vert\, \text{otherwise} }\right.

Hence n 1\mathbb{R}^{n_1} and n 2\mathbb{R}^{n_2} do not have the same relative cohomology unless n 1=n 2n_1 = n_2. Since every homeomorphism induces an isomorphism on (relative) cohomology, they cannot be homeomorphic (using excluded middle here).

Via K-Theory

A proof also follows from the properties of the Adams operations on topological K-theory, see for instance (Wirthmuller 12, p. 46).

References

The first proof is due to Brouwer around 1910.

A proof also follows from the properties of the Adams operations on topological K-theory, see for instance

Revised on August 7, 2017 06:26:49 by Alexis Hazell? (118.209.246.172)