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 $H^0(M, \partial M ; \mathbb{Z}_2) = 0$.
Thanks in advance.
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.