nLab topological invariance of dimension

Idea

A fundamental theorem of topology: for $n_1, n_2 \in \mathbb{N}$ two different natural numbers, $n_1 \neq n_2$, then the Cartesian spaces $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ of dimension $n_1$ and $n_2$, respectively, are not homeomorphic (while for $n_1 = n_2$ 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 $n \in \mathbb{N}$ there are surjective continuous maps $\mathbb{R}^1 \to \mathbb{R}^n$ (the Peano curves), in addition to the obvious injective continuous maps $\mathbb{R}^1 \to \mathbb{R}^n$ for $n \geq 1$ (or indeed $\mathbb{R}^m \to \mathbb{R}^n$ for $m \leq n$), because of which there are also bijective (but discontinuous) maps $\mathbb{R}^m \to \mathbb{R}^n$ for $m, n \geq 1$ (by the Schroeder–Bernstein theorem).

Statement

Theorem

(topological invariance of dimension)

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

More generally:

Theorem

(Brouwer invariance of domain theorem)

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

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

This implies theorem :

Lemma

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

Proof

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

Proof

of theorem from theorem

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

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

Proofs

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

Via ordinary cohomology

Proof

of theorem

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

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

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

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

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

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

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

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

Hence $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ do not have the same relative cohomology unless $n_1 = n_2$. 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).

Related concepts

• Tietze extension theorem

• Brouwer's fixed point theorem

References

The first proof is due to Brouwer around 1910.

• Terry Tao, Brouwer’s fixed point and invariance of domain theorems, and Hilbert’s fifth problem

• Wikipedia, Invariance of domain

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

• Klaus Wirthmüller, p. 46 of Vector bundles and K-theory, 2012 (pdf)