(Theorem) A short exact sequence of chain complexes induces a long exact sequence of homology groups

Overview

The connecting homomorphism $\delta :H_p(C_{\ast },\partial )\to H_{p-1}(A_{\ast },\partial )$ is a construction that associates a short exact sequence of chain complexes

$$0\to (A_{\ast },\partial ){\to }_f(B_{\ast },\partial ){\to }_g(C_{\ast },\partial )\to 0$$

to a long exact sequence of homology groups

$$\cdots {\to }_{\delta }{H}_p(A_{\ast },\partial ){\to }_{f_{\ast }}{H}_p(B_{\ast },\partial ){\to }_{g_{\ast }}{H}_p(C_{\ast },\partial ){\to }_{\delta }{H}_{p-1}(A_{\ast },\partial ){\to }_{f_{\ast }}\cdots .$$

---

Preliminaries: The connecting homomorphism

Short exact sequence of chain complexes

A short exact sequence of chain complexes consists of composable chain maps $f:(A_{\ast },\partial )\to (B_{\ast },\partial )$ and $g:(B_{\ast },\partial )\to (C_{\ast },\partial )$ such that the sequence

$$0\to A_p{\to }_{f_p}{B}_p{\to }_{g_p}{C}_p\to 0$$

is exact for all $p$.

Existence of the connecting homomorphism

Given a short exact sequence of chain complexes, there is a well-defined homomorphism $\delta :H_p(C_{\ast },\partial )\to H_{p-1}(A_{\ast },\partial )$, called the connecting homomorphism, defined by

$$\delta ([c])=[a]$$

when $f(a)=\partial b$ for some $b\in B_p$ with $g(b)=c$.

Sketch of proof from Algebraic Topology.

The proof involves the following key steps:

• Existence of well-defined choices of $b$ and $a$: Let $c\in C_p$ be a cycle (i.e., $\partial c=0$). Surjectivity of $g=g_p:B_p\to C_p$ (due to its location in the SES) implies that we can choose $b\in B_p$ such that $g(b)=c$, and by definition of a chain map we have $g_{p-1}(\partial b)=\partial c=0.$ By exactness, we have $\partial b\in \ker (g_{p-1})=\text{Im}(f_{p-1})$ as well, so there exists an $a\in A_{p-1}$ such that $f(a)=\partial b$.
• Why $a$ is a cycle corresponding to a homotopy class $[a]$: Since $f$ is a chain map, we have $f(\partial a)=\partial f(a)={\partial }^{2}b=0.$ Since $f$ is injective (due to its location in the SES), it has a trivial kernel, so we must have $\partial a=0$ and therefore $a\in A_{p-1}$ is a cycle corresponding to a class $[a]\in H_{p-1}(A_{\ast },\partial )$.
• The class $[a]\in H_{p-1}(A_{\ast },\partial )$ does not depend on choice of lift $b\in B_p$: Suppose we have two different lifts of a representative $c\in C_p$, i.e., $c=g(b)=g(\tilde{b})f(a)=\partial b=\partial \tilde{b}.$ Then $g(b)-g(\tilde{b})=g(b-\tilde{b})=0$, so by exactness there exists $x\in A_p$ such that $f(x)=b-\tilde{b}\in B_p$. Then $\partial b=\partial (\tilde{b}+f(x))=\partial \tilde{b}+\partial f(x)=f(a)+f(\partial x)=f(a+\partial x),$ so we have shown that $\delta ([c])=[a]=[a+\partial x]$. But $a$ differs from $a+\partial x$ by a boundary, so by definition $[a]=[a+\partial x]$.
• The class $[a]$ also does not depend on choice of representative for $[c]$: Suppose we have two representatives for the same homology class $[c]=[\tilde{c}]\in H_p(C_{\ast },\partial )$, so $c=\tilde{c}+\partial y$ for some $y\in C_{p+1}$. Since $g$ is surjective, we can write $y=g(x)$ for some $x\in B_{p+1}$, so $c=g(b)+\partial g(x)=g(b+\partial x),$ which implies that $b+\partial x$ is another lift of $c$. But we showed above that $[a]$ does not depend on choice of lift, and indeed, $\partial (b+\partial x)=\partial b+{\partial }^{2}x=\partial b+0=f(a),$ so this gives the same candidate $[a]$ for $\delta ([c])$.
• The map $\delta$ is indeed a homomorphism (of abelian groups): TBD, or exercise. $\square$

---

Statement and proof

A short exact sequence of chain complexes induces a long exact sequence in homology

Let $f:(A_{\ast },\partial )\to (B_{\ast },\partial )$ and $g:(B_{\ast },\partial )\to (C_{\ast },\partial )$ be composable chain maps forming a short exact sequence of chain complexes, and let $\delta :H_p(C_{\ast },\partial )\to H_{p-1}(A_{\ast },\partial )$ be the connecting homomorphism. Then the sequence

$$\cdots {\to }_{\delta }{H}_p(A_{\ast },\partial ){\to }_{f_{\ast }}{H}_p(B_{\ast },\partial ){\to }_{g_{\ast }}{H}_p(C_{\ast },\partial ){\to }_{\delta }{H}_{p-1}(A_{\ast },\partial ){\to }_{f_{\ast }}\cdots$$

is exact.

Proof from Algebraic Topology.  We make use of the fact that if the composition of two maps is zero, then the image of the first is in the kernel of the second, since the kernel of the composition is the whole domain.

• Exactness at $H_p(A_{\ast },\partial )$: By the definition of $\delta$, given some $[c]\in H_{p+1}(C_{\ast },\partial )$, we have $(f_{\ast }\circ \delta )([c])=f_{\ast }([a])=[f(a)]=[\partial b]=0,$ so $f_{\ast }\circ \delta =0$ is precisely the zero map and $\text{Im}(\delta )\subseteq \ker (f_{\ast })$. Conversely, suppose $f_{\ast }(a)=[0]=[f(a)]$, which means that $f(a)$ is homologous to $0$ and therefore is a boundary in $B_p$, meaning there exists $b\in B_{p+1}$ such that $\partial b=f(a)$. Then $\partial g(b)=g(\partial b)=g(i(a))=0,$ so $g(b)$ is a cycle and we have $\delta ([g(b)])=[a]\in H_p(A_{\ast },\partial )$. Thus, $\ker (f_{\ast })\subseteq \mathrm{Im}(\delta )$ as well.
• Exactness at $H_p(B_{\ast },\partial )$: Since $g\circ f=0$ implies $g_{\ast }\circ f_{\ast }=0$, we know that $\text{Im}(f_{\ast })\subseteq \ker (g_{\ast })$. Conversely, let $[b]\in \ker (g_{\ast })$ be a homology class with a representative cycle $b\in B_p$ that is homologous to $0$ (i.e., $\partial b=0$). Then $g_{\ast }([b])=[g(b)]$ is also homologous to $0$, so $g(b)\in C_p$ is a boundary of some $p+1$ chain, i.e., $g(b)=\partial c^{\prime }$ for $c^{\prime }\in C_{p+1}$. Surjectivity of $g$ implies that there exists $b’\in B_{p+1}$ such that $c’=g(b’)$. Then we have equalities $\partial c’=\partial g(b’)=g(b)$, so $g(b-\partial b^{\prime })=g(b)-g(\partial b^{\prime })=g(b)-\partial g(b^{\prime })=0\in \ker (g).$ By exactness in the chain complexes, this implies $b-\partial b’\in \text{Im}(f)$ as well, so there exists some $a\in A_p$ such that $f(a)=b-\partial b’$. Finally, we show $a$ is a cycle, i.e., $a\in \ker ({\partial }_p:A_p\to A_{p-1})$. Taking the boundary of both sides, we have $\partial f(a)=\partial (b-\partial b^{\prime })=\partial b+{\partial }^{2}b=0,$ which means $f(\partial a)=0$. Then injectivity of $f$ implies $\partial a=0$, and so $a$ is indeed a cycle representing a class $[a]\in H_p(A_{\ast },\partial )$. Thus, $f_{\ast }([a])=[b-\partial b^{\prime }]=[b]-[\partial b^{\prime }]=[b],$ and we conclude that $[b]\in \text{Im}(f_{\ast })$ and $\ker (g_{\ast })\subseteq \text{Im}(f_{\ast })$.
• Exactness at $H_p(C_{\ast },\partial )$: Consider an element $g_{\ast }([b])=[g(b)]\in \mathrm{Im}(g_{\ast }))$, where $[b]\in H_p(B_{\ast },\partial )$ has a representative lift $b\in B_p$. Since $b$ is a cycle, we have $\partial b=0=f(0)$ by injectivity of $f$, so $\delta ([g(b)])=[0]=0$ and $\delta \circ g_{\ast }=0$, implying $\mathrm{Im}(g_{\ast })\subseteq \ker (\partial )$. Conversely, note that if $\delta ([c])=[a]$ for an element of the kernel $[c]\in \ker (\delta )$, then there exists some $a’\in A_p$ such that $\partial a’=a$. Now since $\partial (b-f(a^{\prime }))=\partial b-f(\partial a^{\prime })=\partial b-f(a)=0,$ the element $b-f(a’)$ is a cycle in $H_p(B_{\ast },\partial )$. Finally, we have $j(b-f(a^{\prime }))=g(b)-g(f(a^{\prime }))=g(b)=c,$ so we conclude that $[c]=g_{\ast }([b-f(a^{\prime })])\in \mathrm{Im}(g_{\ast })$ and $\ker (\delta )\subseteq \mathrm{Im}(g_{\ast })$ as well. $\square$

---

Notes

• This is a functor.

---

Code snippets

f: (A_*, \partial) \to (B_*, \partial)

\textup{Im}