Approaching a Cohomology Computation If $U \subseteq \mathbb{R}^2$ is the complement of $d > 0$ points in the plane, and $H^k$ denotes the $k$-th cohomology group, how should I verify that $H^k (U)$ equals $\mathbb{R}$, $\mathbb{R}^d$, or $0$ if $k$ is equal to 0, equal to 1, or $> 1$, respectively? Will the computation play out analogous to a homology computation? As a knee-jerk reaction, I was thinking induction might be of service here, but I am not sure.
Any constructive feedback would be appreciated! 
 A: Your intuition is correct, induction on $d$ is a straightforward way to go
The Mayer-Vietoris sequence for cohomology is
$\dots\rightarrow H^{i}(X)\rightarrow H^i(U)\oplus H^i(V)\rightarrow H^i (U\cap V)\rightarrow H^{i+1}(X)\rightarrow\dots$
By homotopy invariance, the base case follows from knowing the cohomology of the circle (also a Mayer-Vietoris computation).
Since $H^*$ is invariant under homotopy we can arrange that the deleted points are $(1,0),\dots,(d,0)$.  Then apply the cohomology Mayer-Vietoris sequence to the subspaces $U=\{ (x,y)\ |\ x<d \}$ (containing the first $d-1$ deleted points) and $V=\{ (x,y)\ |\ x>d-1 \}$.  We see that $H^i(U\cap V)=0$ for $i>0$ (since $U\cap V$ is contractible) and hence  $H^i(X)\cong H^i(U)\oplus H^i(V)$for all $i>1$ (where $X$ is $\mathbb{R}$ with $d$ points deleted).  This gives the result for $i>1$.  
The rest of the sequence looks like
$0\rightarrow H^0(X)\rightarrow H^0(U)\oplus H^0(V)\rightarrow H^0 (U\cap V)\rightarrow H^1(X)\rightarrow H^1(U)\oplus H^1(V)\rightarrow 0$, so in particular there is a surjection $h:H^1(X)\rightarrow H^1(U)\oplus H^1(V)$.  If we can show that the map $H^0(U)\oplus H^0(V)\rightarrow H^0 (U\cap V)$ is also a surjection (I'm not sure how to do this part right now...), we will know $h$ is an isomorphism, giving the result for $i=1$.  We already know the result for $i=0$ since $H^0(X)=\mathbb{R}$ for any connected space.
