105 reputation
10
bio website
location
age
visits member for 2 years, 6 months
seen Dec 7 at 18:46

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.
May
7
comment Show that $f(x)={x\over2}+c$, where $c$ is a constant
What is $g(x+y) - g(x-y)$?
May
7
comment Show that $f(x)={x\over2}+c$, where $c$ is a constant
Hint: Try taking $g(x) = f(x) - \frac{x}{2}$. Can you write the question in terms of $g$?
May
7
asked Easier proof about suspension of a manifold
Apr
23
awarded  Nice Question
Dec
26
comment The cardinality of $\mathbb{R}/\mathbb Q$
If it is the quotient, in fact, without the axiom of choice, it turns out that $\mathbb{R}/\mathbb{Q}$ can have strictly larger cardinality than $\mathbb{R}$... math.stackexchange.com/a/243549/32178
Nov
24
awarded  Supporter
Nov
24
awarded  Editor
Nov
24
revised What's the envelope curve appear in this graph?
Forgot square roots