Timeline for Does pullback in the category of smooth manifolds always exists?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2015 at 14:16 | vote | accept | Asaf Shachar | ||
| Oct 7, 2015 at 11:04 | history | edited | Simon Henry | CC BY-SA 3.0 |
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| Oct 7, 2015 at 11:03 | comment | added | Simon Henry | For your second comment you are absolutely right, I wanted to emphasize the fact that the pullback and the fiber product are the same as sets, but I will correct this. | |
| Oct 7, 2015 at 11:02 | comment | added | Simon Henry | 1) a point of a manifold $M$ is a morphism from the one point manifold to $M$, as such they obey the universal properties of the pullback and hence are the points of the fiber product. 2) Once you know what the smooth maps to fiber product are, the atlas has to be given by smooth injective maps from open subsets of $\mathbb{R^n}$ to you manifold. In our case, the only such maps are map from $\mathbb{R}$, so our manifold would be one dimensional and it is not possible in a one dimensioncal manifold to have two non constant path that cut into only one point like you have in $0$ here. | |
| Oct 7, 2015 at 11:02 | comment | added | Asaf Shachar | Also, a minor technical point: You wrote 'Moreover a map into the fiber product is smooth if and only if the map to the product is smooth'. I think You should replace fiber product with the pullback object, since they are not necessarily equal (at least as topological spaces, although as sets you claim they are...This is of course the subtle point lurking here, showing the fiber product is not a submanifold is not enough to conclude a pullback does not exists). | |
| Oct 7, 2015 at 10:55 | comment | added | Asaf Shachar | Thanks for your answer. I have two questions: 1) I see why the pullback would have to contain as a set all the points in the fiber product. How can I convince myself that it does not contain any other points? 2) Why such a smooth structure on the union of lines does not exist? (The topology on the pullback can be different from the subspace top' of the product $Y \times Y'$ (like in the weird example in the link...), so in general we 'have freedom' in the topology as well not just in the smooth structure | |
| Oct 7, 2015 at 10:35 | history | answered | Simon Henry | CC BY-SA 3.0 |