How can I exhibit a generator of $H^3(S^3,\mathbb Z_3)$? How can I exhibit a generator for the third cohomology group $H^3(S^3,\mathbb Z_3)$ of the $3$-sphere with coefficients in $\mathbb Z_3$?
 A: Let's construct, for example, an explicit cocycle, viewing $S^3$ as a realization of an abstract simplicial complex and computing cohomology using the ordered complex.
Let us triangulate $S^3$ as the boundary of the $4$-simplex. Let $K$ be the corresponding abstract simplicial complex: we can identify the vertex set of $K$ with $V=\{0,1,2,3,4\}$ and the simplices of $K$ are all the proper subsets of $V$. In particular, the set of $2$- and $3$-simplices are the sers $K^2$ and $K^3$ of $3$- and $4$-element subsets of $V$. The chain complex which arises from $K$ ends with $$\cdots\leftarrow\mathbb ZK^2\overset\partial\longleftarrow\mathbb ZK^3$$ where $\mathbb ZK^2$ is the free abelian group with basis the set $K^2$ and similarly for $\mathbb ZK^3$. The boundary map is given on a basis element $(i_0,i_1,i_2,i_3)\in K^3$ by $$\partial(i_0,i_1,i_2,i_3)=(i_1,i_2,i_3)-(i_0,i_2,i_3)+(i_0,i_1,i_3)-(i_0,i_1,i_2).$$ To compute $H^\bullet(S^3,\mathbb Z_3)$, we have to apply the functor $\hom(\mathord-,\mathbb Z_3)$ to the above chain complex and compute the cohomology of the result. Applying the functor, we get a complex which ends in $$\cdots\rightarrow\hom(\mathbb ZK^2,\mathbb Z_3)\overset{\partial^t}\longrightarrow\hom(\mathbb ZK^3,\mathbb Z_3)$$ with the map $\partial^t$ the transpose of $\partial$, so that for each $\phi:\mathbb ZK^2\to\mathbb Z_3$ we have 
$$
(\partial^t\phi)(i_0,i_1,i_2,i_3)=\phi(i_1,i_2,i_3)-\phi(i_0,i_2,i_3)+\phi(i_0,i_1,i_3)-\phi(i_0,i_1,i_2)
$$
for each $(i_0,i_1,i_2,i_3)\in K^3$. Now $H^3(S^3,\mathbb Z_3)$ is the cokernel of the map $\partial^t$, and since we know it is a $1$-dimensional vector space over the field $\mathbb Z_3$, to describe a generator we need only exhibit any element in $\hom(\mathbb ZK^3,\mathbb Z_3)$ which is not in the image of $\partial^t$.
Now, a computation which can be rather boring or not, depending on how you organize it, shows that for each $\phi:\mathbb ZK^2\to\mathbb Z_3$ we have
\begin{multline}
(\partial^t\phi)(1,2,3,4)
-(\partial^t\phi)(0,2,3,4)
+(\partial^t\phi)(0,1,3,4) \\
-(\partial^t\phi)(0,1,2,4)
+(\partial^t\phi)(0,1,2,3)
= 0
\end{multline}
This implies that if we can find any map $\psi:\mathbb ZK^3\to\mathbb Z_3$ which does not have this property, then it will not belong to the image of $\partial^t$ and therefore its class in $H^3(S^3,\mathbb Z_3)$ will be a generator. 
Now, it is clear that if we define $\psi:\mathbb ZK^3\to\mathbb Z_3$  so that
$$\psi(i_0,i_1,i_2,i_3)=\begin{cases}
1, & \text{if $(i_0,i_1,i_2,i_3)=(0,1,2,3)$;} \\
0, & \text{in any other case}
\end{cases}
$$ the above condition does not hold. Therefore $\psi$ represents a generator.
