Let $X$ be a simplicial abelian group. Let $NX$ be its normalised chain complex denoted

...$\rightarrow NX_{K}$ $\rightarrow$ $NX_{k-1}$ $\rightarrow$...

Now define a new chain complex $Y$ by shifting the normalised chain complex; i.e. $Y_{k} = NX_{k+1}$. Now applying the functor from Dold Kan correspondence on $Y$ we get a simplicial abelian group $\Gamma(Y)$ whose homotopy groups are isomorphic to those of $\Omega X$, since simplical abelian groups are cofibrant-fibrant one can say $\Gamma(Y)$ is isomorphic to $\Omega X$ in homotopy category (using Whitehead's theorem) provided there is a map $\Omega X$ $\rightarrow$ $\Gamma(Y)$ inducing an isomorphism of homotopy groups.

My question is , is there such a canonical map $\Omega X$ $\rightarrow$ $\Gamma(Y)$ .


You can find such a map, but it goes the other way: $\Gamma(Y)\to \Omega X$. It is always wise to keep one's right adjoints on the right hand side.t

But first, let me note that every simplicial abelian group $Z$ is equivalent to $\prod_{k\ge0} K(\pi_kZ,k)$ (this is just a restatement of the classical fact that every chain complex of $\mathbb{Z}$-modules is quasi-isomorphic to its homology), so just the observation about homotopy groups suffices to conclude that $\Gamma(Y)$ and $\Omega X$ are equivalent.

Now, for the construction of the map, note that there is a canonical diagram $$\require{AMScd} \begin{CD} Y @>>> 0\\ @VVV @VVV\\ 0 @>>> N(X) \end{CD} $$ equipped with a chain homotopy of the zero map $Y\to N(X)$ with itself (the latter is just the canonical isomorphism $Y[1]\to N(X)$).

Since Dold-Kan sends chain homotopies to simplicial homotopies, after applying $\Gamma$ we get a homotopy commutative square $$\require{AMScd} \begin{CD} \Gamma(Y) @>>> *\\ @VVV @VVV\\ * @>>> X \end{CD} $$ which again induces the required map from $\Gamma(Y)$ to the homotopy pullback of the rest of the diagram, that is $\Omega X$.

There is a more concrete (although messier) way of obtaining the above map, and it is via constructing the adjoint map $\Sigma \Gamma(Y)\to X$. This is because you can get the suspension via the Kan suspension functor ([1], III.5) $$(\Sigma K)_n = K_n \vee K_{n-1}\vee \cdots \vee K_0$$ so to get map $\Sigma \Gamma(Y)\to X$ you just need to give maps $$(\Gamma(Y))_l=\bigoplus_{[l]\twoheadrightarrow [i]} Y_i = \bigoplus_{[l]\twoheadrightarrow [i]} (NX)_{i+1}\to X_n =\bigoplus_{[n]\twoheadrightarrow [i]} (NX)_{i}$$ for every $l\le n$ satisfying some compatibilities. Spoiler alert: these will just be the canonical inclusions.

In fact what we are doing is just identifying $X$ with $\bar W \Gamma(Y)$, where $\bar W$ is the Eilenberg-MacLane classifying space for a simplicial group ([1] Remark III.5.6).

[1] Goerss, Jardine Simplcial Homotopy Theory, 2006


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