let $S \subseteq \mathbb R^3 \times \mathbb R^3$ be the set of pairs $(x,y)$ where x,y are orthogonal unit vectors in $\mathbb R^3$.

i am trying to show that this is a topological manifold without using smooth maps and other tools of differential topology/ diff. geometry. i am aware that the proof becomes short once one uses the preimage theorem.

definition: a topological manifold is a second countable Hausdorff space that is locally Euclidean.

my idea:

note that $S$ is the inverse image of the inner product of $\{0\}$ and therefore closed. here it says (among other stuff) the following: If the inclusion map $i : S \to M$ is closed then $S$ is actually an embedded submanifold of $M$. "embedded" is unfortunately a notion of differential topology (which i'm trying to avoid here). but i was wondering whether it holds for topological manifolds. therefore:

my question: does it hold that if $M$ is a topological manifold and $S$ is a subset with the property $i : S \to M$ is a closed map then $S$ is a topological submanifold of $M$?

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    $\begingroup$ I don't think this is algebraic topology so much as it is point set topology. Probably the easiest way to do this is to exhibit explicit charts for $S$, which after all is what the preimage theorem is really doing. $\endgroup$ – Zhen Lin Mar 11 '13 at 9:22
  • $\begingroup$ @ZhenLin right, thank you, i retagged accordingly. i will think about what i can do with charts, i don't think it is exactly what the preimage theorem is saying as charts can be homeomorphisms but the map in the theorem is smooth. $\endgroup$ – tom b. Mar 11 '13 at 14:23
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    $\begingroup$ Have you not read the proof of the preimage theorem? It finds explicit charts. $\endgroup$ – Chris Eagle Mar 13 '13 at 19:32
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    $\begingroup$ To answer the question in your bounty, open subsets of manifolds are manifolds, but in general, closed sets need not be. Consider, for example, the Cantor Set. $\endgroup$ – Jason DeVito Mar 13 '13 at 22:11
  • $\begingroup$ Regarding your question, the answer is no. Take a look at a point-set topology textbook in the "uniform convergence" section. More generally it seems like you're fighting-against the formalism of manifolds. I think it's perhaps easy to somehow get the impression that topological manifolds are somehow "more basic" than smooth manifolds, but really the opposite is true. Topological manifolds are a relatively un-natural concept compared with smooth manifolds. Smooth manifolds are objects that are locally linear, and their maps are similarly locally linear. Topl manifolds are complicated. $\endgroup$ – Ryan Budney Mar 19 '13 at 16:01

It is well known that any subset of a Hausdorff space is also Hausdorff with the relative topology. The same occurs with the second countability property. If you need more details, you can read the prove of those facts below.

For the locally Euclidian property, you have to construct the charts of $S$ explicitly. I guess that $\dim S = 3$, since you need two parameters to fix one vector on a sphere and once you fix it you have a 1-parameter family of options to choose the orthogonal (a circle). Hence, for a given point, you could try to find a local homeomorphism from a neighborhood of it to an open set in $S^{2}\times S^{1}$ and, since it is locally Euclidean, then you will be done.

The proof that $S$ is a Hausdorff space and second countable follows from the following facts:

  1. If $A\subset X$ and $(X,\tau)$ is Hausdorff, then $A$ is Hausdorff with the relative topology.

Indeed, take $x,y\in A$ distinct. Then $x,y\in X$. By the Hausdorff axiom of $X$, there exist $U,V\in\tau$ such that $x\in U$, $y\in V$ and $U\cap V=\emptyset$. Then, $U\cap A$ and $V\cap A$ are disjoint open sets in $A$ separating $x$ and $y$.

  1. If $A\subset X$ and $(X,\tau)$ has a basis $\mathcal{B}$, then $$ \mathcal{B}_{A} = \{B\cap A \mid B\in\mathcal{B}\} $$ is a basis of $A$ with the relative topology. In particular, if $X$ is second countable, then $A$ is so.

Indeed, if $W$ is an open set in $A$, then there exists an open set $U$ of $X$ such that $W=U\cap A$. Since $\mathcal{B}$ is basis, then there exists a family $\mathcal{B}'$ of $\mathcal{B}$ such that the union of its elements is $U$: $$ U = \bigcup_{B\in\mathcal{B'}}B. $$ Then, by the distributive property of the intersection, $$ W = \bigcup_{B\in\mathcal{B'}}B\cap A, $$ i.e., $W$ is union of elements of $\mathcal{B}_{A}$.


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