Invariance of domain

Invariance of domain is a theorem in topology about homeomorphic subsets of Euclidean space $\R^n$.
It states:
:If $U$ is an open subset of $\R^n$ and $f : U \rarr \R^n$ is an injective continuous map, then $V := f(U)$ is open in $\R^n$ and $f$ is a homeomorphism between $U$ and $V$.

The theorem and its proof are due to L. E. J. Brouwer, published in 1912.
The proof uses tools of algebraic topology, notably the Brouwer fixed point theorem.

Notes

The conclusion of the theorem can equivalently be formulated as: "$f$ is an open map".

Normally, to check that $f$ is a homeomorphism, one would have to verify that both $f$ and its inverse function $f^$ are continuous;
the theorem says that if the domain is an subset of $\R^n$ and the image is also in $\R^n,$ then continuity of $f^$ is automatic.
Furthermore, the theorem says that if two subsets $U$ and $V$ of $\R^n$ are homeomorphic, and $U$ is open, then $V$ must be open as well.
(Note that $V$ is open as a subset of $\R^n,$ and not just in the subspace topology.
Openness of $V$ in the subspace topology is automatic.)
Both of these statements are not at all obvious and are not generally true if one leaves Euclidean space.

It is of crucial importance that both domain and image of $f$ are contained in Euclidean space .
Consider for instance the map $f : (0, 1) \to \R^2$ defined by $f(t) = (t, 0).$
This map is injective and continuous, the domain is an open subset of $\R$, but the image is not open in $\R^2.$
A more extreme example is the map $g : (-1.1, 1) \to \R^2$ defined by $g(t) = \left(t^2 - 1, t^3 - t\right)$ because here $g$ is injective and continuous but does not even yield a homeomorphism onto its image.

The theorem is also not generally true in infinitely many dimensions. Consider for instance the Banach space $\ell^$ of all bounded real sequences.
Define $f : \ell^\infty \to \ell^\infty$ as the shift $f\left(x_1, x_2, \ldots\right) = \left(0, x_1, x_2, \ldots\right).$
Then $f$ is injective and continuous, the domain is open in $\ell^$, but the image is not.

Consequences

If $n>m$, there exists no continuous injective map $f:U\to\R^m$ for a nonempty open set $U\subseteq\R^n$. To see this, suppose there exists such a map $f.$ Composing $f$ with the standard inclusion of $\R^m$ into $\R^n$ would give a continuous injection from $\R^n$ to itself, but with an image with empty interior in $\R^n$. This would contradict invariance of domain.

In particular, if $n\ne m$, no nonempty open subset of $\R^n$ can be homeomorphic to an open subset of $\R^m$.

And $\R^n$ is not homeomorphic to $\R^m$ if $n\ne m.$

Generalizations

The domain invariance theorem may be generalized to manifolds: if $M$ and $N$ are topological -manifolds without boundary and $f : M \to N$ is a continuous map which is locally one-to-one (meaning that every point in $M$ has a neighborhood such that $f$ restricted to this neighborhood is injective), then $f$ is an open map (meaning that $f(U)$ is open in $N$ whenever $U$ is an open subset of $M$) and a local homeomorphism.

There are also generalizations to certain types of continuous maps from a Banach space to itself. Topologie des espaces abstraits de M. Banach. ''C. R. Acad. Sci. Paris'', 200 (1935) pages 1083–1093

See also

* Open mapping theorem for other conditions that ensure that a given continuous map is open.

Notes

References

*
*
*
*
*
* (see p. 72–73 for Hirsch's proof utilizing non-existence of a differentiable retraction)
*
*
*
*
*
*
*

External links

*

{{Topology

Algebraic topology
Theory of continuous functions
Homeomorphisms
Theorems in topology