invariance of domain

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 $f^\{-1\}$ are continuous;

the theorem says that if the domain is an {{em|open}} subset of $\backslash R^n$ and the image is also in $\backslash R^n,$ then continuity of $f^\{-1\}$ is automatic.

Furthermore, the theorem says that if two subsets $U$ and $V$ of $\backslash R^n$ are homeomorphic, and $U$ is open, then $V$ must be open as well.

(Note that $V$ is open as a subset of $\backslash R^n,$ 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.

File:A map which is not a homeomorphism onto its image.png

It is of crucial importance that both domain and image of $f$ are contained in Euclidean space {{em|of the same dimension}}.

Consider for instance the map $f\; :\; (0,\; 1)\; \backslash to\; \backslash R^2$ defined by $f(t)\; =\; (t,\; 0).$

This map is injective and continuous, the domain is an open subset of $\backslash R$, but the image is not open in $\backslash R^2.$

A more extreme example is the map $g\; :\; (-1.1,\; 1)\; \backslash to\; \backslash R^2$ defined by $g(t)\; =\; \backslash left(t^2\; -\; 1,\; t^3\; -\; t\backslash right)$ 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 lp space $\backslash ell^\{\backslash infty\}$ of all bounded real sequences.

Define $f\; :\; \backslash ell^\backslash infty\; \backslash to\; \backslash ell^\backslash infty$ as the shift $f\backslash left(x\_1,\; x\_2,\; \backslash ldots\backslash right)\; =\; \backslash left(0,\; x\_1,\; x\_2,\; \backslash ldots\backslash right).$

Then $f$ is injective and continuous, the domain is open in $\backslash ell^\{\backslash infty\}$, but the image is not.

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

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

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

The domain invariance theorem may be generalized to manifolds: if $M$ and $N$ are topological {{mvar|n}}-manifolds without boundary and $f\; :\; M\; \backslash to\; N$ 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.{{aut|Leray J.}} Topologie des espaces abstraits de M. Banach. C. R. Acad. Sci. Paris, 200 (1935) pages 1083–1093

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

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Category:Algebraic topology

Category:Theory of continuous functions

Category:Homeomorphisms

Category:Theorems in topology