Normalized chain complex

Given a simplicial abelian group ${\displaystyle A_{\bullet }\in {\text{Ob}}({\text{s}}{\textbf {Ab}})}$ there is a chain complex ${\displaystyle NA_{\bullet }}$ called the normalized chain complex with terms

${\displaystyle NA_{n}=\bigcap _{i=0}^{n-1}\ker(d_{i})\subset A_{n}}$

and differentials given by

${\displaystyle NA_{n}\xrightarrow {(-1)^{n}d_{n}} NA_{n-1}}$

These differentials are well defined because of the simplicial identity

${\displaystyle d_{i}\circ d_{n}=d_{n-1}\circ d_{i}:A_{n}\to A_{n-2}}$

showing the image of ${\displaystyle d_{n}:NA_{n}\to A_{n-1}}$ is in the kernel of each ${\displaystyle d_{i}:NA_{n-1}\to NA_{n-2}}$. This is because the definition of ${\displaystyle NA_{n}}$ gives ${\displaystyle d_{i}(NA_{n})=0}$. Now, composing these differentials gives a commutative diagram

${\displaystyle NA_{n}\xrightarrow {(-1)^{n}d_{n}} NA_{n-1}\xrightarrow {(-1)^{n-1}d_{n-1}} NA_{n-2}}$

and the composition map ${\displaystyle (-1)^{n}(-1)^{n-1}d_{n-1}\circ d_{n}}$. This composition is the zero map because of the simplicial identity

${\displaystyle d_{n-1}\circ d_{n}=d_{n-1}\circ d_{n-1}}$

and the inclusion ${\displaystyle {\text{Im}}(d_{n})\subset NA_{n-1}}$, hence the normalized chain complex is a chain complex in ${\displaystyle {\text{Ch}}_{\geq 0}({\textbf {Ab}})}$. Because a simplicial abelian group is a functor

${\displaystyle A_{\bullet }:{\text{Ord}}\to {\textbf {Ab}}}$

and morphisms ${\displaystyle A_{\bullet }\to B_{\bullet }}$ are given by natural transformations, meaning the maps of the simplicial identities still hold, the normalized chain complex construction is functorial.