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From Wikipedia:

The broadest common definition of manifold is a topological space locally homeomorphic to a topological vector space over the reals.

A topological manifold is a topological space locally homeomorphic to a Euclidean space.

In both concepts, a topological space is homeomorphic to another topological space with richer structure than just topology. On the other hand, the homeomorphic mapping is only in the sense of topology without referring to the richer structure.

I was wondering what purpose it is to map from a set to another with richer structure, while the mapping preserves the less rich structure shared by both domain and codomain? How is the extra structure on the codomain going to be used? Is it to induce the extra structure from the codomain to the domain via the inverse of the mapping? How is the induction like for a manifold and for a topological manifold?

Thanks!

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    Yes, exactly: the extra structure on the codomain is pulled back to give additional local structure on the domain.2011-08-14
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    Thanks! How is the pull back like for a manifold and a topological manifold?2011-08-14
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    I find your question confusing. What else is there to do? If you're going to say some object $X$ is locally like object $Y$, how can you do that in a way that gives $X$ *more* structure than $Y$?2011-08-14
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    @Ryan: That is because I am confused. "If you're going to say some object X is locally like object Y, how can you do that in a way that gives X more structure than Y?" is also my question.2011-08-14
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    Well there's a simple answer to that. If $X$ has all the structure of $Y$, $X = Y$. So this is the most uninteresting case of a manifold.2011-08-14
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    @Dylan: What does "trivial " mean?2011-08-14
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    In some cases, the (local) similarity between the manifold and $\mathbb R^n$ does not extend beyondthe purely topological aspect, as in the graph of f(x)=|x| with the subspace topology; in this case, you can pullback the topology but not the differential part . In other cases, the local similarity does go beyond the topological and into the differentiable, so you can pullback both parts, like, e.g., on the n-sphere, and then you can do calculus on manifolds--by working in $\mathbb R^n$ and using charts to go back to the manifold.2011-08-14
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    Related: http://math.stackexchange.com/questions/53021/defining-a-manifold-without-reference-to-the-reals2011-08-14

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Yes, $\mathbb{R}^n$ has a richer structure than its topology, but we are not interested in those other structures. The topology of $\mathbb{R}^n$ is very special and corresponds to our most intuitive way of thinking of a "space". So, we are only asking that our manifold is a reasonable geometric object.

In other words, when you define a manifold you DO NOT want to use any other structure of an Wuclidian space, just the topological structure. That's why it's defined by a homeomorphism (a map that preserves the topological structure).

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    Thanks! "when you define a manifold you DO NOT want to use any other structure of an euclidian space, just the topological structure." Then if replacing "Euclidean space" with a topological space in the definition of a topological manifold, will the definition still be the same? Similar question for the definition of a manifold, if replacing "a topological vector space over the reals" with a topological space.2011-08-14
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    @Tim: No, the definition will definitely not be the same. A general topological space can have properties that are very different from those of Euclidean space, and the point is that you don't want your manifold to have any such strange topology. It should topologically look like Euclidean space.2011-08-14
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    @Hans: Thanks! Is there another purpose that the domain is endowed with the extra structure of the codomain via the inverse of the mapping?2011-08-14
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    @Tim: I don't think I understand that question, I'm afraid...2011-08-14
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    @Hans: To make an analogy, in measure theory, a mapping from a measure space to a measurable space can induce a measure on the codomain.2011-08-14
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    @Tim: My problem was with parsing the sentence. I just don't understand what is asked, even though I understand each word.2011-08-14
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    @Hans: "you don't want your manifold to have any such strange topology. It should topologically look like Euclidean space." Is Enclidean space the only one that has the type of topology that it has? I.e., is there some other set with the same topology as Euclidean space, but is not an Euclidean space? If yes, can it replace Euclidean space in the definition of topological space?2011-08-16
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    @Tim: Well, there are infinitely many, since any manifold will (locally) have the same type of topology as Euclidean space! But I guess that's not what you're after. ;-)2011-08-16
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    @Hans: I think the definition of topological manifold requires the set that it is locally homeomorphic to globally have the topology of Euclidean space, so another topological manifold cannot replace Euclidean space in the definition.2011-08-16
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    @Tim: Hmm... My first thought was that "locally resembling a sphere" is the same as "locally resembling a plane" ("since it's just local anyway") but perhaps that's not the case?2011-08-16
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The reason to use topological vector spaces as model spaces (for differential manifolds, that is) is that you can define the differential of a curve in a topological vector space. And you can use this to define the differential of curves in your manifold, i.e. do differential geometry.

For more details see my answer here.

All finite dimensional topological vector spaces of dimension n are isomorph to $\mathbb{R}^n$ with its canonical topology, so there is not much choice. But in infinite dimensions things get really interesting :-)

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Part of what is neglected by the seeming presupposition in the last paragraph is that it says "locally". It's only locally the same, not necessarily globally. Thus a sphere or a torus locally looks like a plane, but is not connected together in the same way.

One thing often added to the definitions is that a manifold is a Hausdorff space. This is not redundant. Some manifolds locally homeomorphic to Euclidean spaces are not Hausdorff spaces. For example take a line with one point missing, and then put two points where that one point was. Then define an open neighborhood of either of those two points to contain the point itself plus the other points of some open neighborhood of the missing point. Those two new points cannot be separated from each other by open sets, so it's not a Hausdorff space.