Invariance of domain

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

Normally, to check that ${\displaystyle f}$ is a homeomorphism, one would have to verify that both ${\displaystyle f}$ and its inverse function ${\displaystyle f^{-1}}$ are continuous; the theorem says that if the domain is an open subset of ${\displaystyle \mathbb {R} ^{n}}$ and the image is also in ${\displaystyle \mathbb {R} ^{n},}$ then continuity of ${\displaystyle f^{-1}}$ is automatic. Furthermore, the theorem says that if two subsets ${\displaystyle U}$ and ${\displaystyle V}$ of ${\displaystyle \mathbb {R} ^{n}}$ are homeomorphic, and ${\displaystyle U}$ is open, then ${\displaystyle V}$ must be open as well. (Note that ${\displaystyle V}$ is open as a subset of ${\displaystyle \mathbb {R} ^{n},}$ and not just in the subspace topology. Openness of ${\displaystyle V}$ 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.

A map which is not a homeomorphism onto its image: ${\displaystyle g:(-1.1,1)\to \mathbb {R} ^{2}}$ with ${\displaystyle g(t)=\left(t^{2}-1,t^{3}-t\right).}$

It is of crucial importance that both domain and image of ${\displaystyle f}$ are contained in Euclidean space of the same dimension. Consider for instance the map ${\displaystyle f:(0,1)\to \mathbb {R} ^{2}}$ defined by ${\displaystyle f(t)=(t,0).}$ This map is injective and continuous, the domain is an open subset of ${\displaystyle \mathbb {R} }$, but the image is not open in ${\displaystyle \mathbb {R} ^{2}.}$ A more extreme example is the map ${\displaystyle g:(-1.1,1)\to \mathbb {R} ^{2}}$ defined by ${\displaystyle g(t)=\left(t^{2}-1,t^{3}-t\right)}$ because here ${\displaystyle g}$ 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 L^p space ${\displaystyle \ell ^{\infty }}$ of all bounded real sequences. Define ${\displaystyle f:\ell ^{\infty }\to \ell ^{\infty }}$ as the shift ${\displaystyle f\left(x_{1},x_{2},\ldots \right)=\left(0,x_{1},x_{2},\ldots \right).}$ Then ${\displaystyle f}$ is injective and continuous, the domain is open in ${\displaystyle \ell ^{\infty }}$, but the image is not.

An important consequence of the domain invariance theorem is that ${\displaystyle \mathbb {R} ^{n}}$ cannot be homeomorphic to ${\displaystyle \mathbb {R} ^{m}}$ if ${\displaystyle m\neq n.}$ Indeed, no non-empty open subset of ${\displaystyle \mathbb {R} ^{n}}$ can be homeomorphic to any open subset of ${\displaystyle \mathbb {R} ^{m}}$ in this case.

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

There are also generalizations to certain types of continuous maps from a Banach space to itself.[2]

• Open mapping theorem – Theorem that holomorphic functions on complex domains are open mapsPages displaying wikidata descriptions as a fallback for other conditions that ensure that a given continuous map is open.

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