Reputation
Top tag
Next privilege 250 Rep.
View close votes
Badges
12
Impact
~2k people reached

  • 0 posts edited
  • 0 helpful flags
  • 6 votes cast
Aug
19
comment What is the sheafification of the presheaf of the one point compactification?
Yes, I see that you doubly edited your comment and now I agree with you on all counts. :)
Aug
19
comment What is the sheafification of the presheaf of the one point compactification?
@Eric, I wasn't claiming for this injection to be a homeomorphism (it's obviously not surjective). I was suggesting that maybe $\mathcal{F}^{sh}(U)$ has the subspace topology inherited from this injection (although I don't think this is obvious and I'm not convinced it's true). Also, PyRulez suggested weakening to the category of all topological spaces, which I think is still an interesting question.
Aug
19
comment What is the sheafification of the presheaf of the one point compactification?
@PyRulez: I forgot about the category! It's probably most direct to try to consider the universal property at this point? A first guess might be that the topology comes from being a subset of $\prod_{x\in U}\mathcal{F}_x$ where each $\mathcal{F}_x$ has the topology Cubohr pointed out (and which is now edited into my answer). But I'm not certain...
Aug
19
revised What is the sheafification of the presheaf of the one point compactification?
added 119 characters in body
Aug
19
comment What is the sheafification of the presheaf of the one point compactification?
@Cubohr, you're right about the stalk having $x$ as the only closed point. Which means my general statement about $\mathcal{F}_x$ is also wrong. The strong topology in general will put $\infty$ as the only open point. Will edit accordingly.
Aug
19
answered What is the sheafification of the presheaf of the one point compactification?
Jul
11
awarded  Autobiographer
Jun
30
awarded  Informed
Jun
2
comment Easier proof about suspension of a manifold
I'm not quite sure I see it. Although the cones on $X$ do form a neighbourhood basis for the topology around $p$ and $q$, it is not obvious to me that this means they are necessarily homeomorphic to a disk. Maybe a priori $\Sigma X$ could be a manifold, and a disk around $p$ or $q$ looks kind of funky in the $X \times [0,1]$?
May
16
awarded  Commentator
May
16
comment Nontrivial cycles in the zero set of a map
You're right, the proof is wrong. ($f$ was meant to be $h$, but $\|h(0,0,0,1)+v\|$ will be zero somewhere). In thinking about the problem, I had found that I could show each slice $\{\theta\} \times B^4$ had to have some zero set, but couldn't figure out why these needed to piece together to find a loop which winds around the $S^1$ coordinate once.
May
16
awarded  Benefactor
May
16
comment Easier proof about suspension of a manifold
Awesome. I suppose it's pretty much impossible to get around using Poincaré.
May
16
accepted Easier proof about suspension of a manifold
May
9
answered Nontrivial cycles in the zero set of a map
May
9
awarded  Promoter
May
7
comment Easier proof about suspension of a manifold
Cheers! TeXing was never my strong point.
May
7
comment How many monoids of order three are there?
What monoids are you counting? Maybe you could list them out.
May
7
comment Easier proof about suspension of a manifold
I was more curious if there was some more elementary solution. I figured MSE would be a better place to start for that, but I suppose it could be moved later.
May
7
comment Easier proof about suspension of a manifold
No, there's something a bit more subtle in that last paragraph. In fact, if a homology sphere is not a sphere, then $\Sigma M$ is not a manifold, and so we cannot apply the reasoning of the last paragraph. (See the first comments on the MO question.) The point is that we needed $\Sigma M$ to be a manifold to conclude that it was a sphere, and then for $M$ after that.