Let $\Omega^{n+1}S^{n+1}$ be the base-pointed $(n+1)$-iterated loop space on the $(n+1)$-sphere. In the paper [The homology of $\mathcal{C}_{n+1}$-spaces, $n\geq 0$, F. Cohen, Lecture notes in mathematics Vol. 533,][1] page 226, Theorem 3.2, it is proved that as a $AR_n\Lambda_n$-Hopf algebra,
$$
H_*(\Omega^{n+1}S^{n+1};\mathbb{Z}_p)\cong GW_n H_*(S^0;\mathbb{Z}_p)
$$
for primes $p\geq 2$. 

**Question:** In page 223, it is said that "the coproduct are determined by the diagonal Cartan formulas (page 213, (4))". I do not understand how to write the coproduct of $H_*(\Omega^{n+1}S^{n+1};\mathbb{Z}_p)$ explicitly? 

In the paper [On the homology of configuration spaces,][2] Section 4, as an associative and commutative $\mathbb{Z}_p$-algebra, a generator set of $H_*(\Omega^{n+1}S^{n+1};\mathbb{Z}_p)$ is given. I want to know how to obtain the coproduct formulas of $H_*(\Omega^{n+1}S^{n+1};\mathbb{Z}_p)$ as a coalgebra. 

-------------------------------------------------------------------------------------------------------------------------------------==========================================================================

For the answer below, I have a question.

In the paper [On the homology of configuration spaces][1], Section 4.1, a basis for the graded vector space $H_*(\Omega^m S^m;\mathbb{Z}_2)$ is given as

$(*)$
$$
u_0, u_1, Q_I u_0, Q_I u_1
$$
where $0,1$ denote the two generators of $H_0(S^0;\mathbb{Z}_2)$ and $Q_I=Q_{i_1}Q_{i_2}\cdots Q_{i_r}$ for $I=(i_1,i_2,\cdots, i_r)$,  $0<i_1\leq i_2\leq \cdots\leq i_r<m$. 

By the Cartan formula (cf. [The homology of $\mathcal{C}_{n+1}$-spaces, $n\geq 0$, F. Cohen][2], page 213)
[![enter image description here][3]][3]

In the answer below, we calculated that 

$(**)$
$$\Delta_* Q_I u_0=\sum_{t_1=0}^{i_1}\sum_{t_2=0}^{i_2}\cdots \sum _{t_r=0}^{i_r}Q_{t_1}Q_{t_2}\cdots Q_{t_r}u_0\otimes Q_{i_1-t_1}Q_{i_2-t_2}\cdots Q_{i_r-t_r}u_0,
$$

$(***)$
$$\Delta_* Q_I u_1=\sum_{t_1=0}^{i_1}\sum_{t_2=0}^{i_2}\cdots \sum _{t_r=0}^{i_r}Q_{t_1}Q_{t_2}\cdots Q_{t_r}u_1\otimes Q_{i_1-t_1}Q_{i_2-t_2}\cdots Q_{i_r-t_r}u_1.
$$

Is this result correct or wrong?

**Question:** The problem is that $Q_{t_1}Q_{t_2}\cdots Q_{t_r}u_0$, $Q_{i_1-t_1}Q_{i_2-t_2}\cdots Q_{i_r-t_r}u_0$, $Q_{t_1}Q_{t_2}\cdots Q_{t_r}u_1$, $Q_{i_1-t_1}Q_{i_2-t_2}\cdots Q_{i_r-t_r}u_1$ are not included in the basis. It does not make sense. Why? 

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

I find the Adem relations for $p=2$
[![enter image description here][4]][4]
[![enter image description here][5]][5]

i.e.

for $t_1>2t_2$, 

$$
Q_{t_1}Q_{t_2}=\sum_{t_1/2 \leq i\leq t_1-t_2-1}(-1)^{t_1+i}(2i-t_1,t_1-t_2-i-1)Q^{t_1+t_2-i}Q^i.
$$

How to apply the Adem relation to simplify equations $(**)$, $(***)$ to sums of tensor products of basis given in $(*)$?


  [1]: http://www3.nd.edu/~taylor/papers/cocs.pdf
  [2]: http://www.math.uchicago.edu/~may/BOOKS/homo_iter.pdf
  [3]: https://i.sstatic.net/bAcdw.png
  [4]: https://i.sstatic.net/2BNZy.png
  [5]: https://i.sstatic.net/y2JlL.png