Topological Manifold

A space is a \(d\)-dimensional topological manifold when every one of its points sits inside an open set that can be matched, point for point and continuously in both directions, with an open subset of \(\mathbb{R}^d\). Each point brings its own open set and its own map, and no two of them are asked to agree, so the condition is local: it can hold on a space that as a whole resembles \(\mathbb{R}^d\) in no way at all. One thing is asked of the whole space, and it is the number \(d\), which is the same for every point.

The display is a test to be run at a point. Choose a point, and something has to be produced for it: an open set of \(\mathcal{O}\) holding it, and a map from that set onto its image in \(\mathbb{R}^d\) meeting the four conditions. Nothing requires those sets to overlap, to agree where they do overlap, or to assemble into anything larger.

Conditions (i) to (iii) make \(x\) a homeomorphism between \(U_p\) and \(x(U_p)\), each of those two sets carrying its inherited subset topology. Continuity of \(x^{-1}\) is therefore a statement about the open sets of \(x(U_p)\) in the topology \(x(U_p)\) inherits, and in that topology \(x(U_p)\) is open no matter what set it is, since the whole of a space is always one of its own open sets.

Openness in \(\mathbb{R}^d\) is a stronger demand, and it does not follow. Take \(M = \mathbb{Q}\) with the subset topology inherited from the standard topology on \(\mathbb{R}\), let \(d = 1\), and for every \(p\) take \(U_p = \mathbb{Q}\) with \(x\) the map sending each rational to itself in \(\mathbb{R}\). That \(x\) is invertible and continuous both ways, so (i), (ii) and (iii) hold at every point, yet \(x(U_p) = \mathbb{Q}\) is not open in \(\mathbb{R}\): no soft ball \(B_r(p)\) is made of rationals alone, so \(\mathbb{Q}\) is not a member of \(\mathcal{O}_s\). Condition (iv) is what rules this out, which is why it is stated rather than derived.

What the conditions deliver is a homeomorphism between one open set \(U_p\) and one open subset \(x(U_p)\) of \(\mathbb{R}^d\). It is tempting to promote that to a statement about the whole space and say that a \(d\)-dimensional topological manifold is homeomorphic to \(\mathbb{R}^d\), and it is not true: \(x\) reaches a subset of \(\mathbb{R}^d\) and is asked for nothing beyond it, and outside \(U_p\) the definition says nothing whatsoever.

A wire loop, viewed in \(\mathbb{R}^3\) with the subset topology inherited from the standard topology there, is a \(1\)-dimensional topological manifold. It is not homeomorphic to \(\mathbb{R}\). Take a point out of the loop and what is left is still in one piece; take a point out of \(\mathbb{R}\) and the two half-lines separate what is left. A homeomorphism between the loop and \(\mathbb{R}\) would restrict to one between the leftovers, and it carries open sets to open sets in both directions, so it would carry that separation back to the loop.

Because the quantifier is \(\forall p \in M\), a space is disqualified by a single point at which nothing can be produced. A curve that runs along and then splits into two branches at a point, taken with the subset topology it inherits from \(\mathbb{R}^2\), fails to be a topological manifold because of that splitting point: there is no invertible, both ways continuous map of the kind the definition asks for on any open set holding it.

Suppose \(p\) is the splitting point and some \(U_p\) and \(x\) satisfied the conditions at \(d = 1\). Because \(x(U_p)\) is open in \(\mathbb{R}\), it holds an interval around \(x(p)\), and deleting \(x(p)\) leaves the points near it on two sides. Deleting \(p\) from \(U_p\) leaves them on three, one along each branch. A homeomorphism would have to carry three onto two.