Homeomorphism

A homeomorphism sets up a one-to-one correspondence between the points of \(X\) and \(Y\) that is also a correspondence between their open sets: \(U\) is open in \(X\) if and only if \(f(U)\) is open in \(Y\). Hence everything defined only through open sets (closed sets, interiors, closures, continuity) passes intact from one space to the other. If a homeomorphism exists the spaces are called homeomorphic and we write \(X\cong Y\); this is an equivalence relation, since the identity, inverses and compositions of homeomorphisms are homeomorphisms.

Being bijective and continuous is not enough. The identity from \((\mathbb{R},\text{discrete})\) to \((\mathbb{R},\text{usual})\) is a continuous bijection whose inverse is not, because \(\{0\}\) is open in the first space but not in the second. The requirement on \(f^{-1}\) guarantees that open sets are neither «broken» nor «glued». Later we will see conditions (such as compactness) under which a continuous bijection is a homeomorphism.

A homeomorphism may stretch, bend and twist, but it cannot cut or glue. For example, \(f(x)=\frac{x}{1-x}\) maps \((0,1)\) onto \((0,\infty)\) with inverse \(y\mapsto\frac{y}{1+y}\), so the bounded interval and the half-line are homeomorphic even though one has finite length. This is why distances are not a topological invariant.

To prove two spaces are not homeomorphic, find a topological property that one has and the other lacks (number of open sets, existence of disjoint open sets, and later compactness or connectedness). To prove they are, exhibit the homeomorphism and check the continuity of \(f\) and of \(f^{-1}\). Counting points or open sets is not enough.