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Apr
19
reviewed Approve On the weakly sequential completeness of the dual of the James space $J$
Apr
9
comment When is the torsion submodule a direct factor?
Certainly, if $\mathbf{V}_T/\mathbf{V}_T^\mathrm{tors}$ is free, it is projective, hence the exact sequence $0\to\mathbf{V}_T^\mathrm{tors}\to \mathbf{V}_T\to\mathbf{V}_T/\mathbf{V}_T^\mathrm{tors}\to 0$ splits. From this, you see that an apparently weaker condition is that the quotient be projective, but over a PID projective implies free (see wikipedia, projective module) hence no new modules occur.
Apr
8
comment Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
@TimCampion Ah, yes, I see. Very elegant.
Apr
8
comment Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
@WillSawin I am sure that this is trivial, but how can you conclude, exactly? I mean, once you construct $PGL_2(\overline{\mathbf F}_p)$, how do you get a contradiction assuming there is an equivalence of categories between the two?
Apr
5
awarded  Nice Answer
Apr
5
comment Grothendieck topologies on $\mathbb{C}$
thanks, very nice an complete answer.
Apr
5
accepted Grothendieck topologies on $\mathbb{C}$
Apr
4
revised Grothendieck topologies on $\mathbb{C}$
added 84 characters in body
Apr
4
comment Grothendieck topologies on $\mathbb{C}$
@SimonHenry Oh yes, thanks! I'll correct again; any idea about my initial question about topologies?
Apr
4
comment Numerable covers from the point of view of Grothendieck topologies
ACL Yes, I agree with you. I have spent some time thinking about your remark, and I think you are right. Thanks again.
Apr
4
revised Grothendieck topologies on $\mathbb{C}$
added 10 characters in body
Apr
4
comment Grothendieck topologies on $\mathbb{C}$
Oh yes, sure, thanks. Let me modify my question accordingly— I hope it works. That being said, I would like more to focus on G-tpologies in my question...
Apr
3
asked Grothendieck topologies on $\mathbb{C}$
Apr
1
comment Numerable covers from the point of view of Grothendieck topologies
@ACL One option is that I missed the precise definition of "cover" (for topological spaces): do we allow redundant index sets? I have always thought that index sets can be arbitrary (see e.g. FAC, footnote to page 215): for instance, if $X=U_1 \cup U_2$ is a disconnected space, would you consider a cover $\{V_\nu\}_{\nu\in\mathbb{R}}$ given by $V_\nu=U_1$ if $\nu\leq 0$ and $V_\nu=U_2$ if $\nu >0$ to be finite or uncountable?
Apr
1
comment Numerable covers from the point of view of Grothendieck topologies
@ACL Thanks, I will try to perform the precise computation of the topology attached to the pretopology of numerable covers and will comment back in case I can prove they are equal.
Apr
1
comment Numerable covers from the point of view of Grothendieck topologies
@ACL[cont.] the sieve on each A_n obtained by pull-back of R is a covering sieve of A_n because it comes from the numerable cover {A_i\cap U_n}; but R' was not a covering sieve. I had the same problem when trying to check whether the collection of finite coverings (meaning: the sieves defined as the collections of all opens of all elements of all finite covers) should give rise to a topology, and it does not—I believe. Can you give some details? Thanks.
Apr
1
comment Numerable covers from the point of view of Grothendieck topologies
@ACL Reading your question, I tried to check why does the family of numerable covers define a Grothendieck topology and I got stuck. The problem I encounter is that if you are speaking (as you are probably not) abut the pretopology generated by numerable covers, it seems to me that it would give raise to the "usual" G-topology generated by all covers (by SGA4, Exp. II, 1.1.1). If you are speaking about a topology, I cannot check T2) (ibid.): it seems to me that if UU={U_n} is a numerable cover and AA={A_i} is non-numerable (of an open X), and we associate to them sieves R and R' resp. [cont.]
Dec
27
revised Is there $t\in\operatorname{Gal}(\overline{K}/K)$ s.t. $\operatorname{rank}_{\mathbf{Z}_p}((t-1)E_{p^\infty}(\overline{K}))=1$?
improved formatting
Oct
2
awarded  Yearling
Jun
10
comment Extension class and cup product
I was trying to come up with a precise and detailed proof, but ended up in being stuck. If you take $u\in H^p(X;M'')$ to be $1$ (I guess that by $1$ you mean the neutral element of the cohomology group) you do not have something related to the exact sequence you started with, because $H^p(X;M)=\mathrm{Ext}^p(O_X,M'')\neq \mathrm{Ext}^p(M',M'')$...