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May 21 
awarded  Nice Answer 
Dec 17 
awarded  Yearling 
Dec 17 
awarded  Yearling 
Dec 17 
awarded  Yearling 
Jun 2 
awarded  Good Answer 
Dec 18 
awarded  Yearling 
Sep 16 
comment 
Mathematics of Doodling and the Winding Number
Out of curiosity, we may also look at the following: Suppose the curvature changes sign, but if it does, it stays above 1/r, so the corresponding radius of curvature is greater than r. In this case, we do not need to doodle backwards'. It seems likely to me that the formulas still hold (but the proof by polynomial approximation would not work as polynomial approximation has negative infinite curvature now). It would seem that the direct write the line integral' approach applies.

Aug 3 
comment 
Is every representable map a submersion?
Indeed, I do not see why the proof of Lemma 71 in Metzler's preprint works. Let's look at an example: suppose $M=N=L=\mathbb{R}$ and the maps M>N and L>N are x^2 and y^3, respectively. Isn't the pullback just a line? More explicitly: suppose f and g are differentiable functions such that f^2=g^3. Clearly, f/g is continuous. But actually, it seems that it is differentiable. If so, the pullback does exist, but the induced map $M\times_N L\to M\times L$ is not a smooth embedding. 
Aug 1 
comment 
Picard sheaves for elliptic curves
@Joe Silverman: direct image of a line bundle is not the same as image of some divisor representing it. 
Aug 1 
comment 
Picard sheaves for elliptic curves
Shouldn't the Poincare class also include the diagonal: p^*((o))+q^*((o))Diagonal ? 
Jul 22 
answered  divisors on Abelian varieties 
May 27 
comment 
Derived categories of (coherent) sheaves of modules: exceptional images, gluing, and proper descent?
I am not sure how you expect codimension 2 to help in part 3. Basically, the only advantage is that for a locally free sheaf, the local cohomology will be coherent in cohomological degrees zero and one. However, even for locally free sheaves, you gen noncoherent higher local cohomology. And of course, if your sheaf is not locally free (for instance, torsion), you may have noncoherent cohomology in lower degrees, too. So it would seem there is no recollement for coherent categories, only for quasicoherent ones. 
May 20 
comment 
Dmodules on affine space that are regular at infinity
Come to think of it, is this even true? Consider the Dmodule with one generator f and one relation $\partial f=t f$. (One can write it as $Oe^{t^2/2}$ using YBL's notation.) This $D$module has an irregular singularity at infinity, but its Fourier transform is essentially itself, so it has no singularities outside $0$ and $\infty$. 
May 20 
comment 
Dmodules on affine space that are regular at infinity
This is a great answer, but I am confused by the reason is obviously' sentence... is this supposed to prove the if and only if' claim? If so I would like to see more explanation (to me it seems that some facts about local Fourier transform would be required).

May 17 
answered  Generalized Quotschemes 
May 11 
comment 
Products of Ideal Sheaves and Union of irreducible Subvarieties
I would not say that $Z_1$ and $Z_2$ intersect transversally... (They sums of their tangent spaces is less than the tangent space to $X$.) Are we using different definitions? 
Apr 14 
comment 
Are there any finitely generated artinian modules that are not Noetherian?
Wouldn't this statement reduce to cyclic modules (one generator)? Then, at least if the ring is commutative (I can't figure out whether you are making this assumption), a cyclic module is a free module over a quotient ring, and so noetherian and artinian properties appear to be equivalent. 
Apr 12 
comment 
Direct image of structure sheaf under base change
It's not true in general. Basically, you ask whether $f$ is cohomologically flat (in degree zero), see mathoverflow.net/questions/61289/… For a specific counterexample and some discussion, see mathoverflow.net/questions/56019/… Namely (quoting from Allen Knutson's answer to last question): let $f$ be a flat family of curves degenerating into a curve with an embedded point. 
Apr 1 
comment 
Why don't ideals and quotients work well for categories?
@Chris: The quotient (of an abelian category by a Serre subcategory, or a triangulated category by a thick subcategory) does have a universal property, and it is universal in the class of all functors. 
Mar 18 
comment 
coherent sheaves on affine formal schemes
@Wedhorn: I completely agree that there is a potential confusion. My interpretation was based on the fact that the OP uses two different letters $A$ and $\hat A$ for the rings. 