Coordinate charts for manifold NOT homeomorphic?

60 Views Asked by At

In $\enspace$ Bernard SCHUTZ's Geometrical methods of mathematical physics $\enspace$ a manifolds is defined as a set $M$, such that each element in $M$ has an open neighbourhood which has a continuous, bijective mapping onto an open set of $R^n$ for some $n$.

Beside from not declaring $M$ as a topological space and therefore giving no meaning to an open neighbourhood of some point $p \in M$, why does he only demand the coordinate charts to be continuous and bijective? Other definitions of manifolds demand the charts to be homeomorphisms. Is homeomorphism a direct consequence of continuity and bijectivity? If not, does it even make sense for a chart to not be a homeomorphism?

1

There are 1 best solutions below

1
On BEST ANSWER

I don't know about the book, but you probably shouldn't take it too seriously in terms of mathematical rigor when it has "physics" in its title (not that they are not worth reading for other benefits).

At least, not only it should declare $M$ as a topological space, but also Hausdorff (and probably second countable as well, unless you want to allow examples like the long line). Even so, bijective continuity is not sufficient to deduce homeomorphism: Pick an arbitrary set $S$ with continuum cardinality equipped with the discrete topology, and let $\rho: S\rightarrow R^n$ be an arbitrary bijection. Such a map must be continuous but not homeomorphism. Therefore $S$ as a "manifold" (which should have dimension $0$, if we ignore it's not 2nd countable) can have charts of arbitrary dimension $n$, which is certainly not desired.

We should not consider non-homeomorphism as "charts", but continuous functions that are not charts are frequently used in topology/geometry.