SGA Over 9000: Séminaire d'Geometric Approach

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Balarka Sen

22:17

@Dami $H^0(M; \Bbb Z) = [M, K(\Bbb Z, 0)] = [M, \Bbb N] = \Bbb Z$

Daminark

The last equality is what I'm not sure of

Balarka Sen

Ah, so $M$ is connected right? Any continuous map $M \to \Bbb N$ must then have connected image

So it has to send everything to some fixed integer

there are $\Bbb Z$ many maps of that sort.

none of them homotopic to another

Daminark

Do we not have basepoint considerations?

Balarka Sen

Good point. I think you loosen that if you want to work non-reduced cohomology.

In reduced cohomology $\widetilde{H^0}(M; \Bbb Z) = 0$

Daminark

Ah okay unreduced, aight

Balarka Sen

22:21

All of this would be easier if we worked with the cochain complex from which singular cohomology comes from, like normal people :P

But yeah $H_n(M) \cong \Bbb Z$ is a nontrivial fact.

Daminark

Normal is close to normie

Balarka Sen

Even if you don't use Poincare

A: Möbius strip in non-orientable surface

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