221 reputation
45
bio website
location Oxford, United Kingdom
age 25
visits member for 3 years, 5 months
seen Oct 20 at 16:32

May
4
awarded  Necromancer
May
4
awarded  Yearling
Jun
8
awarded  Constituent
Jun
8
awarded  Caucus
Mar
12
answered Projective varieties are the zeroes of finitely many homogenous polynomials
Feb
11
comment $\operatorname{Func}(J,Ab)$ has enough injectives.
The problem I have is with showing that the h(i) define a natural transformation $h:H\to I$. In the meantime, my professor has written a sketch of solution, I will post the solution here as soon as possible.
Feb
7
awarded  Enthusiast
Jan
31
answered Extension and reduction of the structure group
Jan
30
answered All nondegenerate symmetric bilinear forms are equivalent.
Jan
28
revised $\operatorname{Func}(J,Ab)$ has enough injectives.
added 24 characters in body
Jan
28
comment $\operatorname{Func}(J,Ab)$ has enough injectives.
Thank you for your answer! Unforntunately I do not know enough about category theory or topos theory to understand it.
Jan
28
revised $\operatorname{Func}(J,Ab)$ has enough injectives.
Added: Functor constantly equal to injective abelian group is injective object in Func(J,Ab).
Jan
28
awarded  Student
Jan
28
asked $\operatorname{Func}(J,Ab)$ has enough injectives.
Jan
16
awarded  Supporter
Jan
12
answered Infinitely differentiable function with given zero set?
Jan
10
awarded  Editor
Jan
10
revised Is every embedded submanifold globally a level set?
S has to be closed in M.
Jan
10
comment Is every embedded submanifold globally a level set?
Here, the "Variation on the same theme" is actually a consequence of the first part: if you embed a nonorientable manifold in $\mathbb{R}^n$ (or in any other orientable manifold) then it will have nonorientable normal bundle, so in particular it will have nontrivial normal bundle.
Jan
10
comment Is every embedded submanifold globally a level set?
I don't think this is correct, since $M$ can have nontrivial normal bundle (e.g. $\mathbb{R}P^n\subset\mathbb{R}P^{n+1}$) and in this case it can't be the inverse image of a regular value, despite being compact. I apologize if I am making some mistake.