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Results tagged with at.algebraic-topology
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user 1703
Homotopy theory, homological algebra, algebraic treatments of manifolds.
5
votes
Is the Čech cohomology of an orbifold isomorphic to its singular cohomology?
David C's answer gets this nicely, but another way to intuitively think of it is the following.
Choose a fine enough open cover $\{U_i\}_i$ of $\mathcal{O}$ such that each $U_i$ is isomorphic to $\ma …
- 6,242
5
votes
Does $\mathbb C\mathbb P^\infty$ have a group structure?
This is obviously only a partial answer, concerning $B\mathbb{Z}/2 = \mathbb{RP}^\infty$. In particular, I think that there is a much simpler way to show that it has the structure of an Abelian group. …
- 6,242
1
vote
How to get convinced that there are a lot of 3-manifolds?
This does depend (as others have said) on what you mean by "advanced tools", but what about looking at Seifert-Fibred manifolds? These are ones that are locally products $S^1 \times D^2 / C_n$ for som …
- 6,242
2
votes
maps $\mathbb{S}^{n} \to \mathbb{S}^{n}$
Here is a roundabout solution. The shortest path between two points on a sphere is called a geodesic. Is given by the intersection of the sphere and a plane passing through p, q, and the origin.
Now, …
- 6,242
0
votes
(Homotopy theory) When does the 2 of 3 property not imply 2 of 6?
Edit: This answer only makes sense if the actual question was about morphism $X \to A$ and not $X \to Z$.
Couldn't you just pick a category with two objects $x, y$ with only one invertible morphism $ …
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5
votes
Reference request: Equivariant Topology
This is perhaps a bit of an obvious one, but The Moment Map and Equivariant Cohomology by Atiyah and Bott is a safe bet.
4
votes
How to disjoint two cycles with zero intersection?
Edited:
If you require the $Z_i$ to be connected, then this is not true.
Consider $M = \mathbb{R}^2$ less three points (say $-1, 0, 1$), and let $Z_1$ be a loop around $-1$ and $0$, while $Z_2$ is a …
- 6,242
7
votes
Triviality of finite fiber bundles
This shouldn't be true unless you're using a different definition of fiber bundle than I would expect.
Consider the map $S^1 \to S^1$ which maps $z \mapsto z^2$. This is a fibre bundle whose fiber is …
- 6,242
0
votes
Why is this a local constant sheaf
For a bit of intuition, consider $V = \mathbb{C}^2$ and $M = \mathbb{C}$ with $G = \mathbb{Z}/2$ acting diagonally as $\pm 1$ in both cases.
Then for any open set $U$ not containing the origin in the …
- 6,242
1
vote
What is known about the intersection pairing on H^{mid}?
The bilinear form associated to a connected sum should be the direct sum of the two pairings, yes. Intuitively, classes in M do not intersect classes in N, so they are orthogonal with respect to the p …
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