# Homeomorphism

In general topology, a **homeomorphism** is a map between spaces that preserves all topological properties. Intuitively, given some sort of geometric object, a *topological property* is a property of the object that remains unchanged after the object has been stretched or deformed in some way.

For example, a space \(S\) is called *path-connected* if any two points in \(S\) can be joined by a continuous curve. In symbols, path-connectedness requires that for any \(x\), \(y \in S\), there is a continuous function \(\gamma: [0,1] \to S\) such that \(\gamma(0) = x\) and \(\gamma(1) = y\). Path-connectedness should be considered a topological property, since stretching \(S\) would simply stretch \(\gamma\) as well, and the two points in question would remain connected by a path in the new, stretched space.

Formally, a homeomorphism between two topological spaces \(A\) and \(B\) is a bijection \(f: A \to B\) such that \(f\) is continuous and \(f^{-1} : B \to A\) is also continuous. Then, topological properties are *defined* to be those properties of topological spaces that are preserved under homeomorphism. Recall that continuous maps are essentially those which send points close to one another in the domain to points close to one another in the codomain.

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## Example: Open Intervals Of \(\mathbb{R}\)

For any \(a<b\), the open interval \((a,b) \subset \mathbb{R}\) is homeomorphic to \(\mathbb{R}\).

To show this, first note that \((a,b)\) is homeomorphic to \((0,1)\) via the map \(x \mapsto (x-a)/(b-a)\). This is a homeomorphism since it is a continuous bijection whose inverse, \(x\mapsto (b-a)x + a\), is also continuous.

Thus, it will suffice to exhibit a homeomorphism \((0,1) \to \mathbb{R}\). The map \(f: (0,1) \to \mathbb{R}\) given by \[f(x) = \tan\left(\pi x - \frac{\pi}{2} \right)\] will do the trick, since it is a continuous bijection whose inverse, \(x\mapsto (1/\pi) (\arctan(x) + \pi/2)) \), is also continuous.

## Example: Stereographic Projection

Stereographic projection is an important homeomorphism between the plane \(\mathbb{R}^2\) and the \(2\)-sphere minus a point. The \(2\)-sphere \(S^2\) is the set of points \((x,y,z) \in \mathbb{R}^3\) such that \(x^2 + y^2 + z^2 = 1\). Let \(S^2 \setminus \{N\}\) denote the \(2\)-sphere minus its north pole, i.e. the point \((0,0,1)\).

There exists a homeomorphism \(f: S^2 \setminus \{N\} \to \mathbb{R}^2\), which can be described as follows. First, identify the set \(P = \{(x,y,z) \in \mathbb{R} \, : \, z = 0\}\) with \(\mathbb{R}^2\); the map \(P \to \mathbb{R}^2\) given by \((x,y,0) \mapsto (x,y)\) is a homeomorphism.

For a point \(p\in S^2 \setminus \{N\}\), let \(f(p)\) denote the unique point in \(P\) such that the intersection of the segment \(\overline{N f(p)}\) and \(S^2\) is \(p\). In coordinates, this map is precisely \[f(x,y,z) = \left(\frac{x}{1-z}, \frac{y}{1-z} \right)\]

Via similarly defined maps, one can show that that the \(n\)-sphere minus a point (i.e., \(S^n \setminus \{N\}\)) is homeomorphic to \(\mathbb{R}^n\).

Let \(f_{+}: S^2 - \{N\} \to \mathbb{R}^2\) denote stereographic projection from the north pole of \(S^2\), as defined in the article "Homeomorphism." Similarly, let \(f_{-} : S^2 - \{S\} \to \mathbb{R}^2\) denote stereographic projection from the south pole of \(S^2\), i.e. from the point \(S = (0,0,-1)\).

On \(\mathbb{R}^2\), the function \(g:= f_{+} \circ f_{-}^{-1} \) is well-defined. If \(g(3,4) = (a,b)\), what is \(a+b\)?

## Non-Example: \(\mathbb{R}^2\) Is Not Homeomorphic To \(\mathbb{R}\)

The Cartesian plane \(\mathbb{R}^2\) is not homeomorphic to the real line \(\mathbb{R}^1\). This can be proven by contradiction; suppose there is a homeomorphism \(f: \mathbb{R}^2 \to \mathbb{R}\).

Choose \(p\in \mathbb{R}^2\), and consider \(f(p) \in \mathbb{R}\). If \(f\) is a homeomorphism, then \(f\) restricts to a homeomorphism between \(\mathbb{R}^2 \setminus \{p\}\) and \(\mathbb{R} \setminus \{f(p)\}\).

But \(\mathbb{R}^2 \setminus \{p\}\) is path-connected, while \(\mathbb{R} \setminus \{f(p)\}\) is not! Thus, \(f\) cannot be a homeomorphism.

Let \(\mathbb{R}^2 \setminus \mathbb{Q}^2\) denote the Cartesian plane minus all points whose coordinates are both rational. Similarly, let \(\mathbb{R} \setminus \mathbb{Q}\) denote the irrational numbers. Are the spaces \(\mathbb{R}^2 \setminus \mathbb{Q}^2\) and \(\mathbb{R} \setminus \mathbb{Q}\) homeomorphic?

(**Hint**: Is \(\mathbb{R}^2 \setminus \mathbb{Q}^2\) path-connected?)