bicontinuous function

In the mathematical field of topology, a homeomorphism, topological isomorphism, or bicontinuous function is a bijective and continuous function between topological spaces that has a continuous inverse function. Homeomorphisms are the isomorphisms in the category of topological spaces—that is, they are the mappings that preserve all the topological properties of a given space. Two spaces with a homeomorphism between them are called homeomorphic, and from a topological viewpoint they are the same. The word ''homeomorphism'' comes from the Greek words '' ὅμοιος'' (''homoios'') = similar or same and '' μορφή'' (''morphē'') = shape or form, introduced to mathematics by Henri Poincaré in 1895.

Very roughly speaking, a topological space is a geometric object, and the homeomorphism is a continuous stretching and bending of the object into a new shape. Thus, a square and a circle are homeomorphic to each other, but a sphere and a torus are not. However, this description can be misleading. Some continuous deformations are not homeomorphisms, such as the deformation of a line into a point. Some homeomorphisms are not continuous deformations, such as the homeomorphism between a trefoil knot and a circle.

An often-repeated mathematical joke is that topologists cannot tell the difference between a coffee cup and a donut, since a sufficiently pliable donut could be reshaped to the form of a coffee cup by creating a dimple and progressively enlarging it, while preserving the donut hole in the cup's handle.

Definition

A function $f : X \to Y$ between two topological spaces is a homeomorphism if it has the following properties:

* $f$ is a bijection ( one-to-one and onto),
* $f$ is continuous,
* the inverse function $f^$ is continuous ($f$ is an open mapping).

A homeomorphism is sometimes called a bicontinuous function. If such a function exists, $X$ and $Y$ are homeomorphic. A self-homeomorphism is a homeomorphism from a topological space onto itself. "Being homeomorphic" is an equivalence relation on topological spaces. Its equivalence classes are called homeomorphism classes.

Examples

* The open interval $(a,b)$ is homeomorphic to the real numbers $\mathbf$ for any $a < b$. (In this case, a bicontinuous forward mapping is given by $f(x) = \frac + \frac$ while other such mappings are given by scaled and translated versions of the or functions).
* The unit 2- disc $D^2$ and the unit square in $\mathbf^2$ are homeomorphic; since the unit disc can be deformed into the unit square. An example of a bicontinuous mapping from the square to the disc is, in polar coordinates, $(\rho, \theta) \mapsto \left( \frac, \theta\right)$.
* The graph of a differentiable function is homeomorphic to the domain of the function.
* A differentiable parametrization of a curve is a homeomorphism between the domain of the parametrization and the curve.
* A chart of a manifold is a homeomorphism between an open subset of the manifold and an open subset of a Euclidean space.
* The stereographic projection is a homeomorphism between the unit sphere in $\mathbf^3$ with a single point removed and the set of all points in $\mathbf^2$ (a 2-dimensional plane).
* If $G$ is a topological group, its inversion map $x \mapsto x^$ is a homeomorphism. Also, for any $x \in G$, the left translation $y \mapsto xy$, the right translation $y \mapsto yx$, and the inner automorphism $y \mapsto xyx^$ are homeomorphisms.

Non-examples

* R^''m'' and R^''n'' are not homeomorphic for
* The Euclidean real line is not homeomorphic to the unit circle as a subspace of R^''2'', since the unit circle is compact as a subspace of Euclidean R^''2'' but the real line is not compact.
*The one-dimensional intervals ,1/math> and $]0,1[$ are not homeomorphic because one is compact while the other is not.

Notes

The third requirement, that $f^$ be continuous, is essential. Consider for instance the function $f : [0,2\pi) \to S^1$ (the unit circle in $\mathbf^2$) defined by$f(\phi) = (\cos\phi,\sin\phi)$. This function is bijective and continuous, but not a homeomorphism ($S^1$ is compact but $$f^_is_not_continuous_at_the_point_(1,0),_because_although_f^_maps_(1,0)_to_0,_any_f^_is_not_continuous_at_the_point_(1,0),_because_although_f^_maps_(1,0)_to_0,_any_Neighbourhood_(mathematics)">neighbourhood_

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