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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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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. – Zhen Lin Mar 11 '13 at 9:22
@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. – tom b. Mar 11 '13 at 14:23
Have you not read the proof of the preimage theorem? It finds explicit charts. – Chris Eagle Mar 13 '13 at 19:32
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. – Jason DeVito Mar 13 '13 at 22:11
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. – Ryan Budney Mar 19 '13 at 16:01

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