Discussion of Homotopy Type Theory and Univalent Foundations
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* [HoTT] Does (co)homology detect inhabitation?
@ 2018-11-27 22:22 Michael Shulman
  2018-11-30  5:20 ` [HoTT] " Felix Wellen
                   ` (2 more replies)
  0 siblings, 3 replies; 5+ messages in thread
From: Michael Shulman @ 2018-11-27 22:22 UTC (permalink / raw)
  To: homotopytypetheory

Suppose I have an (unpointed) type X such that (unreduced) H_n(X) or
H^n(X) is nonzero for some n.  In the application I have in mind, this
group is nonzero in a very strong sense, e.g. it has the integers as a
direct summand.  Can I conclude (without using excluded middle) that
||X||?

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* [HoTT] Re: Does (co)homology detect inhabitation?
  2018-11-27 22:22 [HoTT] Does (co)homology detect inhabitation? Michael Shulman
@ 2018-11-30  5:20 ` " Felix Wellen
  2018-11-30 21:54 ` Felix Wellen
  2018-12-04 22:33 ` Ali Caglayan
  2 siblings, 0 replies; 5+ messages in thread
From: Felix Wellen @ 2018-11-30  5:20 UTC (permalink / raw)
  To: Homotopy Type Theory

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I guess the answer is no for cohomology, here is why:

I can sketch examples of cohomologically nontrivial schemes over a base 
field k, without any k-points. Such a scheme is an object X in the oo-topos 
of sheaves on the zariski site and its propositional truncation ||X|| will 
be empty at k (||X||(k)=0). So ||X|| can't be the unit type.

For me, the questions makes already sense for the 0th-cohomology group. 
Don't know if that was included in your question.

So for n=0, we can do the following:
Take k=Q (the rationals) and let P=Spec(Q[X]/(X^2-2)) (=Spec(Q(sqrt(2)))).
Then P has no Q-point, since there is no square root of 2 in Q. Now look at 
the sheaf Z constantly the integers. Its value are

Z(U)={Zariski-locally constant functions U -> Z}

So Z(P)=Z which implies Mor(P,Z)=Z for the external functions and staring 
at it for some time convinced me, that there should also be Z-many distinct 
global sections of

P->Z = ||P->Z||_0 = H^0(P,Z)


Now n=1:
For a scheme having still no Q-points but cohmology in degree 1, I propose 
to use projective space over Q(sqrt(2)) as a Q-scheme.
First, let L:=Spec(Q[X,Y]/(Y^2-2)) (=Spec(Q(sqrt(2))[X]). Then L has still 
no Q-points, for almost the same reason as above.
Now let L\{0}:=Spec(Q[X,Y]/(Y^2-2)_(X)), where "_(X)" denotes the 
localization at the multiplicative system given by everything that is not 
in the ideal generated by X.
And construct projective space (over Q(sqrt(2))) as a pushout:

L\{0} -> L
 |       |
 v       v
 L --->  P(Q(sqrt(2)))

Where we use the inclusion and the inversion after inculsion as span.
P(Q(sqrt(2))) has still no Q-points.
We will look at cohomology with coefficents in GL_1=Spec(Q[X,1/X]). So for 
the first cohomology group, we have

H^1(M,GL_1)=||P(Q(sqrt(2)))->BGL_1||_0

But Mor(P(Q(sqrt(2))),BGL_1)=Pic(P(Q(sqrt(2))))=Z is a known fact. 
Avoiding this reference and using some fishy mix between internal and 
external reasoning, one could also argue:
Use the recursion rule of the pushout (as a HIT) to see that maps 

P(Q(sqrt(2)))->BGL_1

are pairs of maps f,g : L->BGL_1 together with an equality f(x)=g(1/x) for 
all x in L\{0}. So no matter what choice of maps f,g : L -> BGL_1 we can 
make (in reality, there is only one), we get a different map for each 
choice of family of equality that glues them.
Let's choose constant maps for f and g, then a family of equalities 

(x : L\{0}) -> f(x)=g(1/x)

is given by a map 

L\{0} -> GL_1

or

Q[X,1/X] -> Q[X,Y]/(Y^2-2)_(X)

And for each k in Z, we have such a map given by X |-> X^k, and they are 
all different, so Z is a factor of H^1(P(Q(sqrt(2))), GL_1).

Am Dienstag, 27. November 2018 17:22:30 UTC-5 schrieb Michael Shulman:
>
> Suppose I have an (unpointed) type X such that (unreduced) H_n(X) or 
> H^n(X) is nonzero for some n.  In the application I have in mind, this 
> group is nonzero in a very strong sense, e.g. it has the integers as a 
> direct summand.  Can I conclude (without using excluded middle) that 
> ||X||? 
>

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* [HoTT] Re: Does (co)homology detect inhabitation?
  2018-11-27 22:22 [HoTT] Does (co)homology detect inhabitation? Michael Shulman
  2018-11-30  5:20 ` [HoTT] " Felix Wellen
@ 2018-11-30 21:54 ` Felix Wellen
  2018-12-04 22:33 ` Ali Caglayan
  2 siblings, 0 replies; 5+ messages in thread
From: Felix Wellen @ 2018-11-30 21:54 UTC (permalink / raw)
  To: Homotopy Type Theory

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Thierry caught a mistake in my answer, so both my counterexamples actually 
don't work at least for the strong requirement that Z is factor.

The problem is, that the cohomology classes/maps P->Z I use are not 
different over Q.

Am Dienstag, 27. November 2018 17:22:30 UTC-5 schrieb Michael Shulman:
>
> Suppose I have an (unpointed) type X such that (unreduced) H_n(X) or 
> H^n(X) is nonzero for some n.  In the application I have in mind, this 
> group is nonzero in a very strong sense, e.g. it has the integers as a 
> direct summand.  Can I conclude (without using excluded middle) that 
> ||X||? 
>

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* [HoTT] Re: Does (co)homology detect inhabitation?
  2018-11-27 22:22 [HoTT] Does (co)homology detect inhabitation? Michael Shulman
  2018-11-30  5:20 ` [HoTT] " Felix Wellen
  2018-11-30 21:54 ` Felix Wellen
@ 2018-12-04 22:33 ` Ali Caglayan
  2018-12-04 23:15   ` Michael Shulman
  2 siblings, 1 reply; 5+ messages in thread
From: Ali Caglayan @ 2018-12-04 22:33 UTC (permalink / raw)
  To: Homotopy Type Theory

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Can it be done with excluded middle?

On Tuesday, 27 November 2018 22:22:30 UTC, Michael Shulman wrote:
>
> Suppose I have an (unpointed) type X such that (unreduced) H_n(X) or 
> H^n(X) is nonzero for some n.  In the application I have in mind, this 
> group is nonzero in a very strong sense, e.g. it has the integers as a 
> direct summand.  Can I conclude (without using excluded middle) that 
> ||X||? 
>

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* Re: [HoTT] Re: Does (co)homology detect inhabitation?
  2018-12-04 22:33 ` Ali Caglayan
@ 2018-12-04 23:15   ` Michael Shulman
  0 siblings, 0 replies; 5+ messages in thread
From: Michael Shulman @ 2018-12-04 23:15 UTC (permalink / raw)
  To: Ali Caglayan; +Cc: HomotopyTypeTheory

Of course.  WIth excluded middle, either ||X|| or not ||X||, and in
the latter case X is empty too so its homology and cohomology are all
0.
On Tue, Dec 4, 2018 at 2:33 PM Ali Caglayan <alizter@gmail.com> wrote:
>
> Can it be done with excluded middle?
>
> On Tuesday, 27 November 2018 22:22:30 UTC, Michael Shulman wrote:
>>
>> Suppose I have an (unpointed) type X such that (unreduced) H_n(X) or
>> H^n(X) is nonzero for some n.  In the application I have in mind, this
>> group is nonzero in a very strong sense, e.g. it has the integers as a
>> direct summand.  Can I conclude (without using excluded middle) that
>> ||X||?
>
> --
> You received this message because you are subscribed to the Google Groups "Homotopy Type Theory" group.
> To unsubscribe from this group and stop receiving emails from it, send an email to HomotopyTypeTheory+unsubscribe@googlegroups.com.
> For more options, visit https://groups.google.com/d/optout.

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Thread overview: 5+ messages (download: mbox.gz / follow: Atom feed)
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2018-11-27 22:22 [HoTT] Does (co)homology detect inhabitation? Michael Shulman
2018-11-30  5:20 ` [HoTT] " Felix Wellen
2018-11-30 21:54 ` Felix Wellen
2018-12-04 22:33 ` Ali Caglayan
2018-12-04 23:15   ` Michael Shulman

Discussion of Homotopy Type Theory and Univalent Foundations

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