Deligne complex and sheaf of chain complex

Question

Deligne complex $\mathbb{Z}(n)$ and some other complexes can be defined as follows, for example, in this paper, wikipedia and nlab, $$\mathbb{Z}(n):=(0\rightarrow \mathbb{Z}\rightarrow \Omega^0 \rightarrow \Omega^1 \cdots \rightarrow \Omega^{n-1}\rightarrow 0)$$

$$\mathbb{R}(n):=(0\rightarrow \mathbb{R}\rightarrow \Omega^0 \rightarrow \Omega^1 \cdots \rightarrow \Omega^{n-1}\rightarrow 0)$$

$$\ \ \ \Omega^{\bullet}:=(0\rightarrow \Omega^0 \rightarrow \Omega^1 \rightarrow \Omega^2 \cdots \rightarrow \Omega^{n}\rightarrow \cdots)$$

$$\Omega^{\bullet\geq n}:=(0\rightarrow 0 \rightarrow 0 \rightarrow 0 \cdots \rightarrow \Omega^{n}\rightarrow \cdots)$$

$$\Sigma^n\Omega^{n}_{clo}:=(0\rightarrow 0 \rightarrow 0 \rightarrow 0 \cdots \rightarrow \Omega^{n}_{clo}\rightarrow 0)\ $$ where $\mathbb{R}$ stands for locally constant real valued functions. It seems to me there are some relations among these complexes, such as:

1. $\Omega^{\bullet\geq n}\simeq\Sigma^n\Omega^{n}$
2. $\Sigma^n\Omega^{n}_{clo} \simeq \mathbb{R}(n)$
3. $\mathbb{Z}(n)$ is the homotopy pullback of $\Omega^{\bullet\geq n}\rightarrow \Omega^{n} \leftarrow \mathbb{Z}$.

These facts might be somewhat straightforward but I failed to figure out how it works. I would be appreciated for any details provided.

Answer 1

Your statements are not entirely true. The correct ones are:

1. $\Omega^{\bullet\geq n} \simeq \Sigma^n\Omega^n_{\mathrm{cl}}$;
2. $\Sigma^n\Omega^n_{\mathrm{cl}}\simeq \mathbb{R}(n)$;
3. $\mathbb{Z}(n)$ is the homotopy pullback of $\Omega^{\bullet\geq n}\to\Omega^{\bullet}\leftarrow\mathbb{Z}$.

Now let us justify these statements.

1. This follows sort of immediately from the Poincaré lemma. The Poincaré lemma can be interpreted as saying that the sequence $$ 0\to \mathbb{R}\to\Omega^0\to\Omega^1\to\ldots\to\Omega^n\to\Omega^{n+1}\to\ldots $$ of sheaves is exact. Therefore, $\Omega^{\bullet\geq n}$ is exact at all places except at the $n$-th place, and its cohomology there is of course given by $\mathrm{ker}(\Omega^n\to\Omega^{n+1})=\Omega^n_\mathrm{cl}$. This proves statement 1.
2. This also follows from the Poincaré lemma in its interpretation above. Namely, the Poincaré lemma implies that the complex $\mathbb{R}(n)$ is exact at all spots except at the $n$-th place (where $\Omega^{n-1}$ stands). Its cohomology there is given by $\Omega^{n-1}/\mathrm{im}(\Omega^{n-2}\to\Omega^{n-1})$, while $$\Omega^n_\mathrm{cl}=\mathrm{ker}(\Omega^n\to\Omega^{n+1})\cong\mathrm{im}(\Omega^{n-1}\to\Omega^n)\cong \Omega^{n-1}/{\mathrm{ker}(\Omega^{n-1}\to\Omega^n)}\cong\Omega^{n-1}/\mathrm{im}(\Omega^{n-2}\to\Omega^{n-1}),$$ where we again used exactness of the sequence of the Poincaré lemma.
3. Another application of the Poincaré lemma shows that $\Omega^\bullet\simeq\mathbb{R}$ (where $\mathbb{R}$ is seen as a complex concentrated in degree $0$). Combining statements 1. and 2., it thus suffices to show that the square $$ \require{AMScd} \begin{CD} \mathbb{Z}(n) @>>> \mathbb{Z}\\ @VVV @VVV\\ \mathbb{R}(n) @>>> \mathbb{R} \end{CD} $$ with the obvious maps is a homotopy pullback (here, $\mathbb{Z}$ is also a complex concentrated in degree $0$). In the model category of bounded below chain complexes with projective model structure (so that fibrations are degreewise epimorphisms), every object is fibrant, and the map $\mathbb{R}(n)\to\mathbb{R}$ is degreewise an epi as well, and hence a fibration. Therefore, the homotopy pullback of the cospan $\mathbb{R}(n)\to\mathbb{R}\leftarrow\mathbb{Z}$ is given by their strict pullback, and that is by direct inspection given by $\mathbb{Z}(n)$.