# The relative cohomology of the Mobius strip and its boundary.

Let $$M$$ be the Mobius strip with boundary $$\partial M$$. I want to calculate the $$\mathbb{Z}$$ and $$\mathbb{Z}_2$$ relative cohomology of the pair $$(M, \partial M)$$. This pair is a good pair, and so this cohomology is equal to the cohomology of $$M/\partial M$$ which is equal to $$\mathbb{R}\mathbb{P}^2$$. Hence I have found the integral cohomology to be equal to $$\mathbb{Z}, 0, \mathbb{Z}_2$$ in dimensions 0, 1 and 2 respectively, and the $$\mathbb{Z}_2$$ cohomology to be $$\mathbb{Z}_2$$ in dimensions 0, 1, and 2. Is this argument logically sound?

I am confused because when I compute this result using Lefschetz Duality I find that the $$\mathbb{Z}_2$$ cohomology in dimension $$0$$ is $$0$$. Since the Mobius band has $$\mathbb{Z}_2$$-homology groups $$\mathbb{Z}_2$$ in dimensions $$0$$ and $$1$$, and $$0$$ elsewhere, applying Lefschetz duality with the boundary of the Mobius band we should get that $$H^0(M, \partial M ; \mathbb{Z}_2) = 0$$.

You've got one bit wrong. Assuming that $$(X,A)$$ is a good pair (or even weaker: a cofibration) then $$H^n(X, A)$$ is isomorphic to $$H^n(X/A)$$ only for $$n>0$$. In dimension $$0$$ you need reduction. So generally, for any $$n$$ we have that $$H^n(X,A)$$ is isomorphic to the reduced cohomology $$\tilde{H}^n(X/A)$$.
And in your case in dimension $$0$$ the reduced cohomology is $$0$$ since $$M$$ is path connected.