Long Exact Sequences in Homology - Examples

A class in $H_1(\mathbb{Z}[1])$ is represented by an integer $x\in \mathbb{Z}=C_0$. In the cone sequence, $(0,x)$ is a lift of that cycle to degree $1$ of $\mathrm{Cone}(\times m)$.

The cone differential sends $(0,x)$ to $(mx,0)$. Therefore [L1] gives ${\mathrm{\partial }}_1([x])=[mx]\in H_0(\mathbb{Z}[0])=\mathbb{Z}\mathrm{.}$ So the connecting morphism is multiplication by $m$.

The displayed cone sequence is degreewise split by [L1].

The source $H_1(\mathbb{Z}[1])$ and target $H_0(\mathbb{Z}[0])$ are both isomorphic to $\mathbb{Z}$, and [L2] identifies the connecting map with $\pm 1$ on that group. Hence it is nonzero.

Since both source and target are stalk complexes in degree $0$, their only nonzero homology group is $H_0\cong \mathbb{Z}$. Applying [L1] to $\times m$ gives the displayed exact segment.

The map in the middle is multiplication by $m$, so its kernel is $0$ and its cokernel is $\mathbb{Z}\mathrm{/}m$. Exactness identifies these with $H_1(\mathrm{Cone}(\times m))$ and $H_0(\mathrm{Cone}(\times m))$ respectively.

The left and right vertical maps are isomorphisms of stalk complexes, hence quasi-isomorphisms. With the left map also equal to $-1$, the middle vertical map commutes with the canonical inclusion $j_n(y)=(y,0)$ and the projection to $C[1]$, so it is a morphism of short exact sequences; it is a chain map because it changes the sign in degree $0$ exactly as needed to compare the differentials $2$ and $-2$.

Both cone complexes have homology $\mathbb{Z}\mathrm{/}2$ in degree $0$ and $0$ elsewhere, so the middle vertical map induces an isomorphism on homology. This agrees with the prediction of [L1]: once the outer two maps are quasi-isomorphisms, the third must be as well.

The three complexes have cohomology only in degrees $0$ and $1$, and the only nontrivial differential is $d_B^0=\times m$. By [L1], there is a long exact cohomology sequence. The groups immediately before $H^0(A)$ and after $H^1(C)$ are zero, so this long exact sequence collapses to the six displayed terms.

Here $H^0(C)=\mathbb{Z}$ and $H^1(A)=\mathbb{Z}$. The connecting map sends $1\in C^0$ to the class of $m\in A^1$, so ${\mathrm{\partial }}^0$ is multiplication by $m$. This exhibits the boundary as a degree-raising map.

The given sequence is short exact term by term, but [L1] shows that its associated homology sequence contains a nonzero connecting map.

A degreewise short exact sequence of homology groups would have zero connecting map. Since the displayed sequence does not, it is a counterexample to exactness of homology.

The map $\theta$ is a chain map because the top differential is multiplication by $2$ and the bottom one is multiplication by $4$, so $4\cdot 1=2\cdot 2$ on degree $1$. Thus the displayed diagram is a morphism of short exact sequences.

By [L2], the top connecting morphism is multiplication by $2$ and the bottom one is multiplication by $4$. Therefore $(\times 2)\circ {\mathrm{\partial }}_{\mathrm{t}\mathrm{o}\mathrm{p}}={\mathrm{\partial }}_{\mathrm{b}\mathrm{o}\mathrm{t}\mathrm{t}\mathrm{o}\mathrm{m}}\circ 1$ on $H_1(\mathbb{Z}[1])$. This is exactly the commuting square asserted by [L1].

Each relative homology group is the homology of a two-term cone complex with differential multiplication by $m$, $nm$, or $n$. Hence the displayed degree-$0$ groups are the corresponding cokernels and all higher groups vanish.

With only degree-$0$ terms remaining, [L1] collapses to a short exact sequence. The first map is multiplication by $n$ modulo $nm$, and the second is reduction modulo $n$, giving the displayed exact sequence.