Mapping Cones Cylinders and Chain Triangles - Examples

Since the source and target are stalk complexes in degree $0$, [L1] leaves only one nonzero differential, namely $d_1:\mathbb{Z}\to \mathbb{Z},d_1(x)=mx\mathrm{.}$

Because $m\mathrm{\ne }0$, $\ker (d_1)=0$ and $\mathrm{coker}(d_1)=\mathbb{Z}\mathrm{/}m\mathbb{Z}$. Therefore the homology groups are exactly as displayed, and [L2] shows that the cone is acyclic precisely when $m=\pm 1$.

Apply [L1] to the zero map $0:C_{\bullet }\to C_{\bullet }$. This gives the first displayed isomorphism.

Apply [L2] to the identity map $1_C:C_{\bullet }\to C_{\bullet }$. This gives the second displayed conclusion.

By [L1], $\mathrm{Cone}(1_C)$ is contractible, so in particular it is acyclic.

Applying [L2] to the identity map gives that $1_C$ is a quasi-isomorphism.

The displayed source complex is acyclic: multiplication by $2$ is injective, reduction modulo $2$ is surjective, and its kernel is $2\mathbb{Z}$, the image of the first map. If it were contractible, [L1] would make its identity null-homotopic; in degree $0$ that would supply a section $\mathbb{Z}\mathrm{/}2\to \mathbb{Z}$ of the quotient map $\mathbb{Z}\to \mathbb{Z}\mathrm{/}2$, which is impossible. Thus the source complex is acyclic and noncontractible.

By [L2], the cone of the displayed zero map is the shift of the source complex. Because the zero map from an acyclic complex to the zero complex induces isomorphisms on all homology groups, [L4] makes that map a quasi-isomorphism; then [L3] makes its shift a quasi-isomorphism too. Hence the cone is acyclic. If the cone were contractible, applying [L3] with shift $[-1]$ would make the source complex contractible, contradicting step 1.1. Thus this cone is acyclic and noncontractible, so it refutes the displayed statement.

Applying [L1] degreewise leaves the displayed two copies of ${\mathbb{Z}}^2$ in degrees $1$ and $0$; all other terms vanish.

The same construction comes with maps $i$ and $p$, and [L2] identifies $p$ as a chain-homotopy equivalence. So this example is an explicit two-term instance of the general factorization theorem.

By [L1], $H_n(\mathbb{Z},\mathbb{Z};\times m)=H_n(\mathrm{Cone}(\times m))\mathrm{.}$

Substituting the explicit cone homology from [L2] gives the displayed relative homology groups.

Apply [L1] to the composable pair $\times m,\times n$ on $\mathbb{Z}[0]$.

Using [L2], the first summand becomes the two-term complex $0\to \mathbb{Z}\stackrel{n}{\to }\mathbb{Z}\to 0$, while the second summand is contractible. This is exactly the displayed calculation.

By [L1], the cone sequence for $\times 2$ is degreewise split. Its middle term is the complex $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to 0.$

If the sequence split as complexes, the middle term would be isomorphic to $\mathbb{Z}[0]\oplus \mathbb{Z}[1]$, whose homology is $\mathbb{Z}$ in degrees $0$ and $1$ by [L2]. But the middle term from step 1.1 has homology $H_0\cong \mathbb{Z}\mathrm{/}2$ and $H_1=0$. Hence no chain splitting exists, so the displayed sequence is a counterexample.