Cohomology of the sphere $S^{n}$ with coefficients in abelian group $G$ I am doing this with the long exact sequence of the pair $(D^{n}/\partial D^{n})$. 
For $n >0$, using the fact that $(D^{n},\partial D^{n})$ is a good pair we have $H^{i}(D^{n},\partial D^{n};G)\simeq \tilde{H^{i}}(D^n/\partial D^n;G)\simeq\tilde{H^{i}}(S^n;G)$, and the fact that $D^n$ is contractible, a portion of the long exact sequence yields
$0 \leftarrow \tilde{H^{i}}(S^n;G) \leftarrow \tilde{H}^{i-1}(S^{n-1};G)\leftarrow 0$
This gives the isomorphisms $\tilde{H^{i}}(S^n;G) \simeq \tilde{H}^{i-1}(S^{n-1};G)$ for $i >0$. 
I compute $\tilde{H}^{0}(S^{0},G)$ using the fact that $\tilde{H}^{0}(X,G)$ can be described as the group of all functions that are constant on path components modulo the functions that are constant on all of $X$; I denote this space by $F_{X}$. In the case $S^{0}=\{-1,1\}$, the map $\varphi:F_{S^{0}} \rightarrow G$ given by $\varphi(f)=f(1)-f(-1)$ is surjective with kernel the functions constant on $S^{0}$ so will give an isomorphism of $\tilde{H}^{0}(S^{0},G)$ with $G$. 
This gives me $G \simeq \tilde{H}^{0}(S^{0},G) \simeq \tilde{H}^{1}(S^{1},G) \simeq \tilde{H}^{2}(S^{2},G)\simeq ...$

I'm having trouble showing that $H^{k}(S^{n},G)=0$ for all $n \neq k$. It's pretty clear but is there a nice way to write this?

 A: Use long exact sequences and go by induction. You've done the base case. In order to compute cohomology groups for $\mathbb{S}^{n+1}$, use the long exact sequence in cohomology for the good pair $(\mathbb{D}^{n+1},\partial(\mathbb{D}^{n+1})\cong\mathbb{S}^n)$. Starting at $H^{k-1}(\mathbb{S}^{n})$, we get:
\begin{align*}
     \dots\to H^{k-1}(\mathbb{S}^n)\to H^k(\mathbb{S}^{n+1})\to H^k(\mathbb{D}^n)\to H^{k}(\mathbb{S}^n)\to H^{k+1}(\mathbb{S}^{n+1})\to\dots
\end{align*}
Here, we used the fact that since $(\mathbb{D}^{n+1},\mathbb{S}^n)$ is a good pair its cohomology is isomorphic to  that of $\mathbb{D}^{n+1}/\mathbb{S}^n$, which is $\mathbb{S}^{n+1}$. By the inductive hypothesis and the fact that the cohomology of a disk is just that of a point, this gives us the sequence (where $k<n-1$)
\begin{align*}
    \dots\to 0\to H^k(\mathbb{S}^{n+1})\to 0\to 0\to H^{k+1}(\mathbb{S}^{n+1})\to\dots
\end{align*}
So, $H^{k}(\mathbb{S}^{n+1})=0$ as well. The LES for $H^{n}$ is slightly different, but try to write out the LES and the same result should follow.
