Editing Invariance of domain (section)

==Notes==

The conclusion of the theorem can equivalently be formulated as: "<math>f</math> is an [[open map]]".

Normally, to check that <math>f</math> is a homeomorphism, one would have to verify that both <math>f</math> and its [[inverse function]] <math>f^{-1}</math> are continuous; 
the theorem says that if the domain is an {{em|open}} subset of <math>\R^n</math> and the image is also in <math>\R^n,</math> then continuity of <math>f^{-1}</math> is automatic. 
Furthermore, the theorem says that if two subsets <math>U</math> and <math>V</math> of <math>\R^n</math> are homeomorphic, and <math>U</math> is open, then <math>V</math> must be open as well. 
(Note that <math>V</math> is open as a subset of <math>\R^n,</math> and not just in the subspace topology. 
Openness of <math>V</math> 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|thumb|alt=Not a homeomorphism onto its image|An injective map which is not a homeomorphism onto its image: <math>g : (-1.1, 1) \to \R^2</math> with <math>g(t) = \left(t^2 - 1, t^3 - t\right).</math>]]
It is of crucial importance that both [[Domain of a function|domain]] and [[Image of a function|image]] of <math>f</math> are contained in Euclidean space {{em|of the same dimension}}. 
Consider for instance the map <math>f : (0, 1) \to \R^2</math> defined by <math>f(t) = (t, 0).</math> 
This map is injective and continuous, the domain is an open subset of <math>\R</math>, but the image is not open in <math>\R^2.</math> 
A more extreme example is the map <math>g : (-1.1, 1) \to \R^2</math> defined by <math>g(t) = \left(t^2 - 1, t^3 - t\right)</math> because here <math>g</math> 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|Banach]] [[lp space|{{mvar|L<sup>p</sup>}} space]] <math>\ell^{\infty}</math> of all bounded real [[sequence]]s. 
Define <math>f : \ell^\infty \to \ell^\infty</math> as the shift <math>f\left(x_1, x_2, \ldots\right) = \left(0, x_1, x_2, \ldots\right).</math>  
Then <math>f</math> is injective and continuous, the domain is open in <math>\ell^{\infty}</math>, but the image is not.