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.

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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