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. ∎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.