3

I have to compute $H^{i}(\mathbb{R}P^2,\mathbb{Z}_2)$. I know that is $\mathbb{Z}_2$ for $i=0,1,2$ but I'm looking for a proof without universal coefficient theorem. Have you some ideas?

ArthurStuart
  • 4,932
  • What is the definition of cohomology group? Do you know about $\operatorname{Hom}$? – Sigur May 19 '13 at 14:59
  • @ No.. I'm sorry – ArthurStuart May 19 '13 at 15:00
  • 2
    What kind of cohomology are you using? Singular, CW, simplicial...? Although they all give the same result in the end, you should tell us which to use to solve your problem so that it may be of use to you. Also, do you wish to know the algebra structure or only the group structure? – Olivier Bégassat May 19 '13 at 15:02
  • @OlivierBégassat I wish to know only the group structure and I know only CW and singular cohomology... – ArthurStuart May 19 '13 at 15:05

1 Answers1

5

If you look at Hatcher page 144 he writes down the cellular chain complex for $\Bbb{R}P^n$ for both even and odd $n$. Consider the case now when $n$ is even and our coefficient theory is in $\Bbb{Z}/2\Bbb{Z}$. Then your cellular chain complex is looking like

$$0 \to \Bbb{Z}/2\Bbb{Z} \stackrel{2\cdot}{\longrightarrow} \Bbb{Z}/2\Bbb{Z} \stackrel{0}{\longrightarrow} \ldots \stackrel{0}{\longrightarrow} \Bbb{Z}/2\Bbb{Z} \to 0.$$

However because our chain groups are now $\Bbb{Z}/2\Bbb{Z}$ the multiplication by two map is the zero map. If you now apply the functor $\hom(-,\Bbb{Z}/2\Bbb{Z})$ you will get the exact same chain complex but with arrows going the other way. Now what happens when you take the homology of a chain complex where all the maps are the zero maps?