How to compute $H^1(\Sigma_g-\{p\})$ using Mayer-Vietoris? How can I find, using Mayer-Vietoris, $H^1(\Sigma_g-\{p\})$, where $\Sigma_g$ is a genus $g$ surface?
 A: The space $\Sigma_g - \{p\}$ deformation-retracts onto a wedge of $2g$ circles, so it suffices to compute cohomology of $\vee_{2g}\mathbb{S}^1$. This can be done quickly by hand, but since the exercise asks you to use Mayer-Vietoris, here's a sketch how to do it in the case of a punctured torus. 
Let $X = \mathbb{S}^1\vee\mathbb{S}^1$, $A$ and $B$ each be one of the circles, and $A\cap B$ be the wedge point. Use Mayer-Vietoris to relate $H^i(A)$, $H^i(B)$, and $H^1(A\cap B)$ to $H^i(X) \cong H^i(\Sigma_g)$; the Mayer-Vietoris long exact sequence for cohomology is
$$0\to H^0(X)\to H^0(A)\oplus H^0(B)\to H^0(A\cap B)\to H^1(X)\to H^1(A)\oplus H^1(B)\to H^1(A\cap B)$$
The basic idea will be to exploit that the map $H^1(X)\to H^1(A)\oplus H^1(B)$ is an epimorphism.
Having computed $H^1(\Sigma_1-\{p\})$, proceed by induction: $\Sigma_g-\{p\}$ deformation retracts onto a wedge of $2g$ circles, which is a bouquet of $2g-2$ circles with two more wedged on. So let $A = \vee_{2g-2}\mathbb{S}^1 \cong (\Sigma_{g-1}-\{p\})$ and $B = \mathbb{S}^1\vee\mathbb{S}^1\cong (\Sigma_1-\{p\})$ and use Mayer-Vietoris again.

Just for fun, here's how to do it by hand. Since $\pi_1$ of a wedge of $n$ circles is free on $n$ generators and $H_1$ is the abelianization of $\pi_1$, we know that $H_1$ of a wedge of $n$ circles is isomorphic to $\mathbb{Z}^n$. Use the relationship between $H^1$ and $\operatorname{Hom}(H_1,\mathbb{Z})$ to find $H^1$.
