[HoTT] Re: Does (co)homology detect inhabitation?

[Felix Wellen <felix.wellen@gmail.com>]

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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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