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19h
asked functor from complex algebraic variety to constructible function
2d
awarded  Popular Question
Dec
1
comment Proving that $n$-cycle has order $n$
That cycle has order n and I showed it has order at least n.
Nov
22
comment Show that solutions of $a^3+b^3=c^3+d^3$ and $a+b=c+d$ for distinct $a,b,c,d$ are non-existent.
You mean positive integer, right?
Nov
21
answered Let three signed measures such that $\lambda_{1}\bot\mu$ and $\lambda_{2}\bot\mu$, then $\left(\lambda_{1}+\lambda_{2}\right)\bot\mu$?
Nov
21
comment Vector bundles on $M/G$
One special case is that when the action of $G$ on $M$ is free, $E\rightarrow M/G$ is a vector bundle with fiber $V$.
Nov
21
comment Evaluating $\int^\infty _{-\infty} \frac{e^{-i p x / h}}{x^2 + a^2}\,\mathrm{d}x $
I think you can find the value by consider the contour integral over a semi-circle living in $\{y\ge 0\}$ with radius $r$ and let $r\rightarrow \infty$
Nov
21
comment Show that there only two solutions to $p \mid (k^2 - 1)$
It is just sth minor. I mean both cases can happen at the same time. In this situation, it turns out that both cases actually are the same.
Nov
21
comment Show that there only two solutions to $p \mid (k^2 - 1)$
the prime $p$ has to be greater than $2$.
Nov
21
comment Confusion about pulling back homogeneous forms.
You mean $\mathbb{P}^2$
Nov
21
answered Compute sum with generating functions
Nov
21
comment How many ways to put 3 balls in 5 urns
Are your balls identical to each other?
Nov
21
comment Infinite closed subset of $S^1$ such that the squaring map is a bijection?
My point is that this problem cannot be solved by considering measure since it will 'neglect' set of zero measure...
Nov
21
comment Infinite closed subset of $S^1$ such that the squaring map is a bijection?
what if the closed set has zero measure?
Nov
21
awarded  Custodian
Nov
21
reviewed Approve Does $f\colon X\to X$ with $f^{-1}(U)=U$ imply $f$ is homeomorphism?
Nov
21
answered Does $f\colon X\to X$ with $f^{-1}(U)=U$ imply $f$ is homeomorphism?
Nov
21
asked preimage of singular locus in resolution of singularities
Nov
11
revised Prove that the natural map $\alpha : \text{Hom}(X,\text{Spec} A) \rightarrow \text{Hom}(A,\Gamma(X,\mathcal{O}_X))$ is an isomorphism
corrected latex
Nov
11
suggested approved edit on Prove that the natural map $\alpha : \text{Hom}(X,\text{Spec} A) \rightarrow \text{Hom}(A,\Gamma(X,\mathcal{O}_X))$ is an isomorphism