Recent content by Jamma

So it's the gradient flow just "going upwards" but multiplied by a scalar? Just drew a picture, and I think I see how it goes, you at least get to a very small retract of the figure 8.

I'm thinking of a torus - how does the flow continue past the first critical point?

Hmm, I may have been wrong. Some browsing reveals this: All manifolds are homotopy equivalent to a CW complex [--interestingly, there is kind of an inverse too: all countable CW complexes of dim n are homotopy equivalent to a differentiable manifold of dimension 2n+1...

I meant with just one puncture. Working on intuition, removing one of the cells at the boundary one by one, you shouldn't change the homotopy type at any stage. The justification for this is that if a cell sits on the boundary, we should be able to deformation retract it to the boundary it...

This seems to be the case - if we at least assume our manifold is a CW complex with contractible n-cells, it seems to be that we can nibble away our n-cells touching the boundary one by one until none are left and at each stage we will have a space of the same homotopy type. I know this isn't a...

I think you are right - given the standard picture of a Klein Bottle, RP2 or torus as a square with identifications, one can "stretch the puncture" out to the boundary so that you have a graph. Thinking about a 3d analogue with a cube with sides identified, it would seem that you can make the...

Nice proof, that wraps it up. Thanks Lavinia! In fact, doesn't this prove the general statement that deformation retracts onto proper subspaces don't exist for all manifolds - I assume that punctured manifolds will always be homotopy equivalent to graphs? You can then use the above argument...

Of course, another way you can easily see this is to just think that the square is the image of the line, but where overlapping points on the line are the same.

Hmm, how about you glue together any two points x,y whenever f(x)=f(y)? The map f applied to this quotient space will then be a bijection. The universal property of quotients will mean that the induced map is continuous and of course a continuous bijection from a compact space to a Hausdorff...

cf. the above with the simple description of the generator of the zeroeth cohomology group for a path connected space - the 0-cochain assigning the value 1 to each 0-chain. It is easy to see that capping with this class will just be the identity, as desired.

Hello all, I'm trying to get my head around the cap product and singular cohomology. I've always found singular cochains rather hard to visualise (e.g. what is the fundamental cochain of a manifold? i.e. that chain which generates the Z in the top cohomology group), and I've found looking at...

I agree. A deformation retract is a htpy equivalence and it feels as though any proper subset of a n-dim manifold should have trivial homology in degree n (Z/2Z coefficients). Could this be made explicit? You need to use closed subsets to have retracts on Hausdorff spaces (I think) and it will...

Well my thinking above was to use the fundamental class of the homology of the manifold. What I said indeed proves that no deformation to a subspace will exist as long as the subspace you have has a trivial H_2 (or, indeed, an H_2 with no 2 torsion) - this includes deformation retracts as a...

This is a very fundamental example. Strictly what you are saying isn't quite correct - this quotient space still is a manifold, since it is locally homeomorphic to the disc everywhere (even the cone point!). Usually in orbifold theory though, you keep track of these cone points, so you have...

What do you mean by saying "not deformable to any of its proper subspaces"? Do you mean a map from RP2 to a subspace of it (f, say?) such that f followed by the inclusion is homotopic to the identity? Couldn't you, for example, prove that if you are retracting to a subspace with no homology...