# 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.

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.
• Thank you for the answer. I have one question (for now): What does the $\times 2$ mean above the $\mathbb{Z_2}\rightarrow \mathbb{Z_2}$ at the 3d paragraph. Also, just bellow you have written $\mathbb{Z_2}/2\mathbb{Z_2}\cong Z_2$. This is $\mathbb{Z_2}$ or something else ? – Nick A. Mar 17 '18 at 15:45