At the page 163 of the book Simplicial Homotopy Theory by Goerss and Jardine, the first sentence starts with "Observe that $N_0A= A$, that ...". According to the further calculation in that page and the original definition at the page before, it looks like $N_0A_0 = A_0$ however $N_0A_n=kerd^n_0 \neq A_n$. Am I missing something? Or is this just a typo with a quick fix that I can't see?
Definition given in the book is as follows: $A_n$ is any simplicial abelian group and $$N_jA_n = \begin{cases} \bigcap^j_{i=0} \text{ker$(d_i)$} & \text{for $ n \geq j+2$,} \\ NA_n & \text{for $n \leq j+1$.} \end{cases} $$
Here $NA_n$ is the n-th component of the normalized chain complex, so $NA_n = \bigcap^{n-1}_{i=0}\text{$d_i$} \subset A_n $, and maps $d_i$ have the domain $A_n$.
Thank you in advance!
After solving this exercise sheet about the normalized chain complex, it seems like the definition is given for $N_{j+1}A_n$ in the book. Which makes $N_0A = A$ indeed. So it should have been: $$N_jA_n = \begin{cases} \bigcap^{j-1}_{i=0} \text{ker$(d_i)$} & \text{for $ n \geq j+1$,} \\ NA_n & \text{for $n \leq j$.} \end{cases} $$ Similar indexing typo is made throughout the construction of the chain homotopy that would prove Theorem 2.4. But the main idea is completely same.