Long Exact Sequences in Homology

Apply [L3] to the commutative square in degree $n+1$. Using [L2], its cokernel row is exactly ${\bar{A}}_n\to {\bar{B}}_n\to {\bar{C}}_n\to 0,$ so the top row is exact at ${\bar{B}}_n$ and ${\bar{C}}_n$.

Apply [L3] to the commutative square in degree $n-1$. By [L2], the resulting kernel row is $0\to Z_{n-1}(A)\to Z_{n-1}(B)\to Z_{n-1}(C),$ exact at $Z_{n-1}(A)$ and $Z_{n-1}(B)$. The two rows are connected by the differentials, and the chain-map equalities from [L1] make the diagram commutative.

Because $q_n$ is the cokernel of ${\beta }_n$, one has $q_n{\beta }_n=0$.

Using [L1], ${\tilde{\mathrm{\partial }}}_n{\beta }_n={\delta }_n^{\mathrm{s}\mathrm{n}\mathrm{a}\mathrm{k}\mathrm{e}}{q}_n{\beta }_n=0.$ Thus the composite from the boundary subobject to $H_{n-1}(A)$ is the zero morphism.

Because $c$ is a cycle, $p_{n-1}(d_n^{B}b)=d_n^C(p_{n}b)=d_n^C(c)=0.$ By [L2] and [L3], $d_n^{B}b$ lies in the image of $i_{n-1}$, so there is a unique $a\in A_{n-1}$ with $i_{n-1}(a)=d_n^{B}b$. The definition in [L1] then gives ${\mathrm{\partial }}_n([c])=[a]$.

If $b^{\mathrm{\prime }}$ is another lift of the same cycle $c$, then $b-b^{\mathrm{\prime }}=i_n(u)$ for some $u\in A_n$ by [L2] and [L3]. Hence $i_{n-1}(a-a^{\mathrm{\prime }})=d_n^B(b-b^{\mathrm{\prime }})=d_n^B(i_n(u))=i_{n-1}(d_n^{A}u)\mathrm{.}$ Since $i_{n-1}$ is injective, $a-a^{\mathrm{\prime }}=d_n^{A}u$ is a boundary, so $[a]=[a^{\mathrm{\prime }}]$.

If $c^{\mathrm{\prime }}=c+d_{n+1}^C(v)$ is another cycle representative of $[c]$, choose a lift $v^{\mathrm{\prime }}\in B_{n+1}$ of $v$. Then $b+d_{n+1}^B(v^{\mathrm{\prime }})$ lifts $c^{\mathrm{\prime }}$, and its boundary differs from $d_n^B(b)$ by $d_n^B{d}_{n+1}^B(v^{\mathrm{\prime }})=0.$ Equivalently, the corresponding element of $A_{n-1}$ differs from $a$ by a boundary. So the class from step 1.1 depends only on $[c]$.

The three maps in [L1] are exactly the maps appearing in the statement.

Exactness of that categorical segment gives $\mathrm{im}({\mathrm{\partial }}_{n+1})=\ker (H_n(A)\to H_n(B)),$ which is the desired equality.

The three maps in [L1] are exactly the maps appearing in the statement.

Exactness of that categorical segment gives $\mathrm{im}(H_n(A)\to H_n(B))=\ker (H_n(B)\to H_n(C)),$ as required.

The three maps in [L1] are exactly the maps appearing in the statement.

Exactness of that categorical segment gives $\mathrm{im}(H_n(B)\to H_n(C))=\ker ({\mathrm{\partial }}_n),$ as required.

The three maps in [L1] are exactly the maps appearing in the statement.

Exactness of that categorical segment gives $\mathrm{im}({\mathrm{\partial }}_n)=\ker (H_{n-1}(A)\to H_{n-1}(B)),$ as required.

The maps $H_n(A)\to H_n(B)$ and $H_n(B)\to H_n(C)$ are defined by [L1], and the connecting map ${\mathrm{\partial }}_n$ is defined by The connecting morphism in homology. Thus the displayed sequence exists in every degree.

For each integer $n$, [L2] gives exactness at $H_n(A)$, $H_n(B)$, $H_n(C)$, and $H_{n-1}(A)$. Therefore every four-term window around ${\mathrm{\partial }}_n$ is exact.

Since $n$ was arbitrary, these exact windows concatenate to the displayed bi-infinite exact sequence.

By [L1], each degree of the given ladder is a morphism of short exact sequences. Passing to the quotient-kernel diagrams used on this page therefore gives a morphism between two weaker Stacks snake diagrams. Regard every comparison arrow as an object of the arrow category. By [L2], the resulting diagram is itself weaker Stacks snake data in that abelian category.

Apply [L4] in the arrow category to the diagram from step 1.1. Its connecting morphism is an arrow object whose two components are the connecting morphisms of the original weaker snake diagrams; being a morphism in the arrow category says exactly that the square between those components commutes. Under the identifications in [L3], this is the displayed homology connecting square.

Each ordinary square in the two long exact sequences is induced by one of the three chain maps in the given ladder, so it commutes by [L1].

The only nonformal squares are the connecting ones, and they commute by [L2]. Therefore every square in the long exact ladder commutes.

By [L1], the given short exact sequence of cochain complexes is the same data as a short exact sequence of reindexed chain complexes. Applying [L3] gives a long exact sequence in homology for those chain complexes.

Replace each term $H_{-n}(X^{\mathrm{♯}})$ by $H^n(X)$ using [L2]. Under the same reindexing, the connecting map from degree $-n$ homology to degree $-n-1$ homology becomes a map ${\mathrm{\partial }}^n:H^n(C)\to H^{n+1}(A)\mathrm{.}$ This is the displayed long exact cohomology sequence.

Apply [L1] to both rows of the given morphism. This turns the cochain ladder into a morphism of short exact sequences of chain complexes.

The resulting lower-index connecting square commutes by [L2]. Translating back with [L3] gives the upper-index square in the statement.

By [L1], the degree-$n$ window is $H_n(B)\to H_n(C)\stackrel{{\mathrm{\partial }}_n}{\to }{H}_{n-1}(A)\to H_{n-1}(B)\mathrm{.}$

The outer terms in that window are zero by [L2]. Exactness then forces ${\mathrm{\partial }}_n$ to be both monic and epic, hence an isomorphism.

If $A_{\bullet }$ and $C_{\bullet }$ are acyclic, then for every $n$ the exact window $H_n(A)\to H_n(B)\to H_n(C)$ has zero outer terms by [L2]. Exactness from [L1] therefore gives $H_n(B)=0$ for all $n$.

If $A_{\bullet }$ and $B_{\bullet }$ are acyclic, then the exact window $H_n(B)\to H_n(C)\to H_{n-1}(A)$ has zero outer terms, so $H_n(C)=0$ for all $n$.

If $B_{\bullet }$ and $C_{\bullet }$ are acyclic, then the exact window $H_n(C)\to H_{n-1}(A)\to H_{n-1}(B)$ has zero outer terms, so $H_{n-1}(A)=0$ for all $n$. Thus the remaining complex is acyclic in every case.

Assume $b$ and $c$ are quasi-isomorphisms. In the long exact ladder from [L1], center the five-term window at $H_n(A)$. The four surrounding comparison maps come from $H_{n+1}(b)$, $H_{n+1}(c)$, $H_n(b)$, and $H_n(c)$, so they are isomorphisms by [L3]. Hence $H_n(a)$ is an isomorphism by [L2].

Assume $a$ and $c$ are quasi-isomorphisms. Center the five-term window at $H_n(B)$. The surrounding comparison maps come from $H_{n+1}(c)$, $H_n(a)$, $H_n(c)$, and $H_{n-1}(a)$, so [L2] gives that $H_n(b)$ is an isomorphism.

Assume $a$ and $b$ are quasi-isomorphisms. Center the five-term window at $H_n(C)$. The surrounding comparison maps come from $H_n(a)$, $H_n(b)$, $H_{n-1}(a)$, and $H_{n-1}(b)$, so [L2] yields that $H_n(c)$ is an isomorphism. By [L3], the missing map is therefore a quasi-isomorphism in every case.

Suppose first that there is a chain section $s$ with $ps=1_C$. Functoriality of homology gives $H_n(p)H_n(s)=1_{H_n(C)}$, so $H_n(p)$ is epic. Exactness in [L3] gives $\ker ({\mathrm{\partial }}_n)=\mathrm{im}(H_n(p))=H_n(C)$, hence ${\mathrm{\partial }}_n=0$.

Suppose instead that there is a chain retraction $r$ with $ri=1_A$. Then $H_{n-1}(r)H_{n-1}(i)=1_{H_{n-1}(A)}$, so $H_{n-1}(i)$ is monic. Exactness in [L3] gives $\mathrm{im}({\mathrm{\partial }}_n)=\ker (H_{n-1}(i))=0$, and a morphism with zero image in an abelian category is zero. Thus the connecting morphism vanishes in either chain-split situation.

Apply [L4] to the short exact sequence from [L1]. This gives an exact sequence $\cdots \to H_n(D)\to H_n(\mathrm{Cone}(f))\to H_n(C[1])\stackrel{{\delta }_n}{\to }{H}_{n-1}(D)\to \cdots \mathrm{.}$

Apply the weaker snake construction [L6] to the quotient-kernel diagram of the cone sequence from [L1] in degree $n+1$. Let $z_C:Z_n(C)\to C_n$ be the cycle inclusion and let $s:Z_n(C)\to \mathrm{Cone}(f{)}_{n+1}$ have components $(0,z_C)$. Its projection to $C[1{]}_{n+1}=C_n$ is $z_C$, so $s$ and the quotient $q_C:Z_n(C)\to H_n(C)$ induce a morphism $t:Z_n(C)\to P$ into the pullback used in [L6], with $\pi t=q_C$. By [L5], $d^{\mathrm{Cone}(f)}s=j_n{f}_n{z}_C,$ and the defining equation for $r$ in the snake construction therefore gives $rt=Z_n(f)$. Consequently [L6] yields ${\delta }_{n+1}{q}_C={\delta }_{n+1}\pi t=q_{D}rt=q_D{Z}_n(f)=H_n(f)q_C\mathrm{.}$ The last equality is the defining square for [L3]. Since $q_C$ is epic, ${\delta }_{n+1}=H_n(f)$. Reindexing step 1.1 by [L2] gives the displayed cone long exact sequence.

A class in $H_n(C[1])$ is represented by a cycle $x\in C_{n-1}$. In the canonical cone sequence, the element $(0,x)\in \mathrm{Cone}(f{)}_n$ lifts that class, and its boundary is $(f_{n-1}(x),0)\mathrm{.}$ By [L2], the connecting morphism sends $[x]$ to $[f_{n-1}(x)]\in H_{n-1}(D)$.

Step 1.1 is exactly the formula for $H_{n-1}(f)$ after identifying $H_n(C[1])$ with $H_{n-1}(C)$ via [L3]. Hence the connecting map of the cone sequence agrees with $H_{n-1}(f)$ under that shift identification. When $f=1_C$, this becomes the shifted identity with precisely the sign encoded in [L3].

If $\mathrm{Cone}(f)$ is acyclic, then the middle terms in [L1] vanish by [L2]. Exactness therefore makes every $H_n(f)$ both monic and epic, so $f$ is a quasi-isomorphism.

If $f$ is a quasi-isomorphism, then every map $H_n(f)$ in [L1] is an isomorphism. Exactness forces each $H_n(\mathrm{Cone}(f))$ to vanish, so the cone is acyclic by [L2]. This is the same criterion already established in [L3].

Apply [L2] to the induced chain map $\alpha :\mathrm{Cone}(f)\to \mathrm{Cone}(gf)$. This gives an exact sequence $\cdots \to H_n(\mathrm{Cone}(f))\to H_n(\mathrm{Cone}(gf))\to H_n(\mathrm{Cone}(\alpha ))\to H_{n-1}(\mathrm{Cone}(f))\to \cdots \mathrm{.}$

Rewrite the first two terms by [L1]. By [L3] there is a chain-homotopy equivalence $\mathrm{\Phi }:\mathrm{Cone}(\alpha )\to \mathrm{Cone}(g)$; [L4] makes $\mathrm{\Phi }$ a quasi-isomorphism, and [L5] therefore gives isomorphisms on homology. So $H_n(\mathrm{Cone}(\alpha ))\cong H_n(\mathrm{Cone}(g))$, and rewriting the last term with [L1] yields exactly the displayed long exact sequence of relative homology for the composable pair.

In the cone long exact sequence from [L1], every copy of $H_n(C)$ and $H_n(D)$ is zero by [L2].

Exactness then forces every $H_n(\mathrm{Cone}(f))$ to vanish as well. Hence the cone is acyclic.

By [L1], applying $F$ degreewise to the given short exact sequence of complexes produces another short exact sequence of complexes in $\mathcal{B}$. Applying [L4] to both sequences gives two long exact homology sequences.

The canonical isomorphisms from [L2] identify each term $F(H_n(X))$ with $H_n(F(X))$, and their naturality identifies the ordinary homology maps on the two sequences.

The connecting morphisms are built from kernels and cokernels of the same degreewise diagram, and [L1] preserves those constructions. Therefore the comparison isomorphisms respect the connecting maps as well, so the whole long exact sequence is transported from one side to the other.

By [L1] and [L2], the long exact homology sequence breaks in each degree into a short exact sequence $0\to H_n(A)\to H_n(B)\to H_n(C)\to 0.$

Because the short exact sequence is chain split, the middle complex is degreewise isomorphic to $A_{\bullet }\oplus C_{\bullet }$ with block-diagonal differential. Hence $Z_n(B)\cong Z_n(A)\oplus Z_n(C),B_n(B)\cong B_n(A)\oplus B_n(C),$ and quotienting cycles by boundaries gives $H_n(B)\cong H_n(A)\oplus H_n(C)\mathrm{.}$

The hypothesis is exactly one of the three cases covered by [L1].

Therefore the middle vertical map is a quasi-isomorphism.

By [L1], the short exact sequence gives a long exact sequence running through the six displayed terms.

Every homology term immediately before and after those six terms is zero by the concentration hypothesis. Removing those zero terms leaves the displayed six-term exact sequence.