If $M^n$ is a compact topological manifold (not necessarily with additional structure), is there an upper bound on the number of charts needed to cover $M$ ? Does this bound depend on the dimension of $M$ ?

Thanks in advance...


  • $\begingroup$ a related thread $\endgroup$ – t.b. Dec 20 '11 at 6:42
  • $\begingroup$ I have seen a related thread about my question. But it seems the manifolds they talk about are triangulated manifolds and i would like to know an answer that does not depend on triangulability of manifolds. And i would like to have a good reference. Thanks in advance !! $\endgroup$ – onebengaltiger Dec 20 '11 at 19:05
  • $\begingroup$ I have examined the related thread a bit more, and it seems i could find some answers by looking at Ostrand's theorem and Kirby-Siebenmann handle decompositions for TOP manifolds. Thanks for the tip about the related thread !!! $\endgroup$ – onebengaltiger Dec 22 '11 at 5:11
  • $\begingroup$ Someone's trying to edit the question to remove the word "compact"... wtf? how is that a reasonable edit?? $\endgroup$ – 6005 Oct 21 '16 at 14:31

This is an old question, but maybe somebody still cares.

First, let us define what a "chart" means in the context of topological manifolds $M^n$:

Definition. A chart on $M$ is an open subset $U\subset R^n$ and a topological embedding $f: U\to M^n$. We say that $M$ is covered by charts $(U_i, f_i), i\in J$, if $$ M=\bigcup_{i\in J} f_i(U_i). $$

Note that in this definition I do not require the open sets $U$ to be connected, which is important. I also require $M$ to be Hausdorff and 2nd countable.


Theorem. Every topological $n$-dimensional manifold $M$ (compact or not) admits a cover by $n+1$ charts.

Proof. Note that $M$ has topological dimension $n$ and is a normal topological space (since $M$ embeds in some $R^N$ and, hence, is metrizable). Let ${\mathcal W}$ be an open cover of $M$ by subsets homeomorphic to open balls in $R^n$. Thus, by Ostrand's theorem on colored dimension, there exists a set $\{{\mathcal V}_i: i=0,...,n\}$ such that:

  1. Elements of each ${\mathcal V}_i$ are certain pairwise disjoint open subsets of $M$.

  2. The union $$ {\mathcal V}=\bigcup_{i=0}^{n} {\mathcal V}_i $$ is an open cover of $M$.

  3. Each element of ${\mathcal V}$ is contained in an element of ${\mathcal W}$.

By 2nd countability property of $M$, each ${\mathcal V}_i$ is at most countable and each of its elements is homeomorphic to an open subset of $R^n$ (part 3). Therefore, for each $i$, the (disjoint!) union $$ T_i=\bigcup_{V\in {\mathcal V}_i} V $$ admits a topological embedding $g_i: T_i\to U_i=g_i(T_i)\subset R^n$. Its inverse $f_i$ is a chart on $M$. Since ${\mathcal V}$ is a cover of $M$, it follows that we obtained a cover of $M$ by $n+1$ charts $(U_i, f_i), i=0,...,n$. qed


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