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Homeomorphism: When Two Spaces Are the Same

If continuous maps are the morphisms of topology, then the isomorphisms are the homeomorphisms — continuous bijections with continuous inverse. They are the precise meaning of the joke that a topologist cannot tell a doughnut from a coffee mug. We pin down the definition, watch a deceptive non-example, and meet the topological invariants that prove two spaces are genuinely different.

The definition, and why the inverse matters

A homeomorphism is a bijection f : X → Y such that both f and f⁻¹ are continuous. When one exists we call X and Y homeomorphic: they are the same topological space wearing different labels. Because f is a continuous bijection with continuous inverse, U is open in X if and only if f(U) is open in Y — so f sets up a perfect dictionary between the two topologies. Every topological property of X is shared by Y.

A worked homeomorphism

Distances and lengths are not topological, so a bounded interval can be homeomorphic to an unbounded line. Here is the cleanest example: the open interval (-1, 1) is homeomorphic to all of R. Topologically a finite open interval and the entire real line are indistinguishable, even though one has length 2 and the other is infinite.

Claim: f : (-1, 1) -> R, f(x) = x / (1 - x^2), is a homeomorphism.

1. Well-defined: for -1 < x < 1 the denominator 1 - x^2 > 0, so f(x) is finite.

2. f is continuous: it is a quotient of polynomials with nonzero denominator
   on (-1,1), hence continuous there.

3. f is strictly increasing, so injective:
   f'(x) = (1 + x^2) / (1 - x^2)^2 > 0 for all x in (-1, 1).

4. f is surjective onto R:
   as x -> 1^-,  f(x) -> +infinity;   as x -> -1^+,  f(x) -> -infinity.
   f is continuous, so by the intermediate value theorem it hits every
   real value in between. Thus f((-1,1)) = R.

5. f^{-1} is continuous:
   solving y = x/(1 - x^2) gives  x = (sqrt(1 + 4y^2) - 1) / (2y)  for y != 0,
   and x = 0 for y = 0; this is continuous on all of R.
   (Equivalently: a continuous strictly monotone bijection on an interval
    automatically has a continuous inverse.)

f is a continuous bijection with continuous inverse  =>  homeomorphism.   ∎
A bounded open interval is homeomorphic to the whole real line — length is not a topological invariant.

Invariants: how to prove spaces differ

To show two spaces are homeomorphic you exhibit a homeomorphism. To show they are not is harder: you must rule out all possible maps. The tool is a topological invariant — a property preserved by homeomorphism. If X has it and Y does not, no homeomorphism can exist. Compactness and connectedness, the subjects of the rest of this track, are the two workhorse invariants.