Homology with coefficients from homology. My main goal is to understand the computations behind the cohomology ring of $\mathbb{C}P^n$ as done in Bott & Tu. To this ends, I am reading a set of notes about Spectral Sequences (here) by Antonio Díaz Ramos. While in general I find them very good and detailed I have some trouble understanding the how the aurthor is forming (in page 6) the second page $E^{2}_{p,q}=H_p(S^2;H_q(S^1;\mathbb{Z}))$. 
What does it mean for a homology to take coefficients from another homology?
I would be very grateful If someone could explain in detail how get $E_2$ of $C_2$ in page 9.
 A: If $G$ is an Abelian group then the singular
homology group $H_p(X;G)$ is defined. It's the homology of $S_p(X)\otimes G$. Here $H_q(S^1;\Bbb Z)$ is the usual singular homology, i.e., $\Bbb Z$
for $q\in\{0,1\}$ and zero otherwise. Thus $E_{p,q}^2$ is zero if $q\ge2$,
otherwise it's $H_p(S^2;\Bbb Z)$, so $\Bbb Z$ for $p\in\{0,2\}$ and
zero otherwise. This is stated on page 6.
On page 9, we have group homology. These are groups $H_p(G;A)$
where $G$ is a group, and $A$ an Abelian group with an action
of $G$. In this example, $G$ always acts trivially.
First of all $H_q(C_2;\Bbb Z)$ is $\Bbb Z$ for $q=0$, $\Bbb Z/2\Bbb Z=
\Bbb Z_2$ for odd $q$ and $\ker(\Bbb Z\stackrel{\times 2}{\longrightarrow}\Bbb Z)=0$ for even $q>0$.
So $E_{p,q}^2=0$ when $q>0$ is even. When $q$ is odd, it is
$H_p(C_2;\Bbb Z_2)$ which is
$\Bbb Z_2$ for $p=0$, $\Bbb Z_2/2\Bbb Z_2\cong Z_2$ for $p$
odd and $\ker(\Bbb Z_2\stackrel{\times 2}{\longrightarrow}\Bbb Z_2)
\cong\Bbb Z_2$ for even $p>0$. Finally $E_{p,0}^2=H_p(C_2;\Bbb Z)$
which we've already worked out. This gives the first diagram on page 9.
