[proofplan] We prove exactness directly from the definition of the mapping cone. First we verify that $i$ and $p$ are chain maps, so that they induce maps on homology. Then we compute the kernel of each displayed homology map: the cone boundary formula shows that $\ker H_n(p)=\operatorname{im} H_n(i)$, that $\ker H_{n-1}(f)=\operatorname{im} H_n(p)$, and that $\ker H_{n-1}(i)=\operatorname{im} H_{n-1}(f)$. Since these three identities hold in every degree, they assemble into the long exact sequence. [/proofplan] [step:Verify that the inclusion and projection are chain maps] For each $n \in \mathbb{Z}$, define the $R$-linear maps \begin{align*} i_n: D_n &\to \operatorname{Cone}(f)_n, & y &\mapsto (y,0), \end{align*} and \begin{align*} p_n: \operatorname{Cone}(f)_n &\to C_{n-1}, & (y,x) &\mapsto x. \end{align*} The shifted complex $C[-1]$ has degree-$n$ term $C_{n-1}$ and differential $-\partial^C_{n-1}: C_{n-1}\to C_{n-2}$. Thus $p$ is a map $\operatorname{Cone}(f)\to C[-1]$. For $y\in D_n$, \begin{align*} \partial^{\operatorname{Cone}}_n i_n(y) &= \partial^{\operatorname{Cone}}_n(y,0)\\ &= \bigl(\partial^D_n y + f_{n-1}(0),-\partial^C_{n-1}0\bigr)\\ &= \bigl(\partial^D_n y,0\bigr)\\ &= i_{n-1}(\partial^D_n y). \end{align*} Therefore $i: D\to \operatorname{Cone}(f)$ is a chain map. For $(y,x)\in \operatorname{Cone}(f)_n$, we compute \begin{align*} p_{n-1}\partial^{\operatorname{Cone}}_n(y,x) &= p_{n-1}\bigl(\partial^D_n y+f_{n-1}(x),-\partial^C_{n-1}x\bigr)\\ &= -\partial^C_{n-1}x. \end{align*} This is exactly the differential of $C[-1]$ applied to $p_n(y,x)=x$. Hence $p:\operatorname{Cone}(f)\to C[-1]$ is a chain map. Consequently $i$ and $p$ induce homology maps \begin{align*} H_n(i):H_n(D)&\to H_n(\operatorname{Cone}(f)),\\ H_n(p):H_n(\operatorname{Cone}(f))&\to H_n(C[-1])\cong H_{n-1}(C). \end{align*} Under the identification $H_n(C[-1])\cong H_{n-1}(C)$, the map $H_n(p)$ sends a cone homology class $[(y,x)]$ to $[x]\in H_{n-1}(C)$. [guided] We first check that the maps in the displayed sequence are actually defined on homology. The inclusion map is the degreewise $R$-[linear map](/page/Linear%20Map) \begin{align*} i_n: D_n &\to \operatorname{Cone}(f)_n, & y &\mapsto (y,0). \end{align*} Using the cone differential, for every $y\in D_n$ we have \begin{align*} \partial^{\operatorname{Cone}}_n i_n(y) &= \partial^{\operatorname{Cone}}_n(y,0)\\ &= \bigl(\partial^D_n y+f_{n-1}(0),-\partial^C_{n-1}0\bigr)\\ &= \bigl(\partial^D_n y,0\bigr)\\ &= i_{n-1}(\partial^D_n y). \end{align*} So $i$ commutes with differentials and is a chain map. The projection is slightly subtler because it lands in a shifted complex. Define \begin{align*} p_n: \operatorname{Cone}(f)_n &\to C_{n-1}, & (y,x) &\mapsto x. \end{align*} The target is $C[-1]$, whose degree-$n$ group is $C_{n-1}$ and whose degree-$n$ differential is $-\partial^C_{n-1}$. Therefore, to prove that $p$ is a chain map, we must check that projecting after taking the cone boundary gives $-\partial^C$ on the $C$-component. Indeed, \begin{align*} p_{n-1}\partial^{\operatorname{Cone}}_n(y,x) &= p_{n-1}\bigl(\partial^D_n y+f_{n-1}(x),-\partial^C_{n-1}x\bigr)\\ &= -\partial^C_{n-1}x. \end{align*} This equals the shifted differential applied to $p_n(y,x)=x$. Hence $p:\operatorname{Cone}(f)\to C[-1]$ is a chain map. Since chain maps send cycles to cycles and boundaries to boundaries, both $i$ and $p$ induce homology maps. Finally, because $H_n(C[-1])$ is naturally identified with $H_{n-1}(C)$, the map $H_n(p)$ sends a class represented by a cone cycle $(y,x)\in D_n\oplus C_{n-1}$ to the class $[x]\in H_{n-1}(C)$. [/guided] [/step] [step:Identify the kernel of $H_n(p)$ with the image of $H_n(i)$] Let $[(y,x)]\in H_n(\operatorname{Cone}(f))$, represented by a cone cycle $(y,x)\in D_n\oplus C_{n-1}$. The cycle condition is \begin{align*} \partial^D_n y + f_{n-1}(x) &= 0,\\ \partial^C_{n-1}x &= 0. \end{align*} First suppose $[(y,x)]\in \operatorname{im} H_n(i)$. Then there is a cycle $z\in D_n$ such that $[(y,x)]=[(z,0)]$, so \begin{align*} H_n(p)([(y,x)])=H_n(p)([(z,0)])=[0]=0. \end{align*} Thus $\operatorname{im}H_n(i)\subseteq \ker H_n(p)$. Conversely, suppose $[(y,x)]\in \ker H_n(p)$. Then $[x]=0$ in $H_{n-1}(C)$, so there exists $a\in C_n$ such that $x=\partial^C_n a$. Define \begin{align*} z := y+f_n(a)\in D_n. \end{align*} Because $f$ is a chain map, \begin{align*} \partial^D_n z &= \partial^D_n y+\partial^D_n f_n(a)\\ &= \partial^D_n y+f_{n-1}(\partial^C_n a)\\ &= \partial^D_n y+f_{n-1}(x)\\ &=0. \end{align*} Thus $z$ is a cycle in $D_n$. Also \begin{align*} \partial^{\operatorname{Cone}}_{n+1}(0,-a) &= \bigl(\partial^D_{n+1}0+f_n(-a),-\partial^C_n(-a)\bigr)\\ &= \bigl(-f_n(a),\partial^C_n a\bigr)\\ &= \bigl(-f_n(a),x\bigr). \end{align*} Therefore \begin{align*} (y,x)-\partial^{\operatorname{Cone}}_{n+1}(0,-a) &= (y,x)-\bigl(-f_n(a),x\bigr)\\ &= (y+f_n(a),0)\\ &= i_n(z). \end{align*} So $[(y,x)]=H_n(i)([z])$, and $\ker H_n(p)\subseteq \operatorname{im}H_n(i)$. Hence \begin{align*} \ker H_n(p)=\operatorname{im}H_n(i). \end{align*} [/step] [step:Identify the kernel of $H_{n-1}(f)$ with the image of $H_n(p)$] Let $[x]\in H_{n-1}(C)$, represented by a cycle $x\in C_{n-1}$ with $\partial^C_{n-1}x=0$. First suppose $[x]\in \operatorname{im}H_n(p)$. Then there is a cone cycle $(y,x)\in \operatorname{Cone}(f)_n$ representing a class whose image under $H_n(p)$ is $[x]$. The cone cycle condition gives \begin{align*} \partial^D_n y+f_{n-1}(x)=0. \end{align*} Thus $f_{n-1}(x)=-\partial^D_n y$ is a boundary in $D_{n-1}$, and therefore \begin{align*} H_{n-1}(f)([x])=[f_{n-1}(x)]=0. \end{align*} So $\operatorname{im}H_n(p)\subseteq \ker H_{n-1}(f)$. Conversely, suppose $[x]\in \ker H_{n-1}(f)$. Then $f_{n-1}(x)$ is a boundary in $D_{n-1}$, so there exists $y\in D_n$ such that \begin{align*} f_{n-1}(x)=-\partial^D_n y. \end{align*} Since $\partial^C_{n-1}x=0$, the pair $(y,x)\in D_n\oplus C_{n-1}$ satisfies \begin{align*} \partial^{\operatorname{Cone}}_n(y,x) &= \bigl(\partial^D_n y+f_{n-1}(x),-\partial^C_{n-1}x\bigr)\\ &=(0,0). \end{align*} Thus $(y,x)$ is a cone cycle, and \begin{align*} H_n(p)([(y,x)])=[x]. \end{align*} Hence $\ker H_{n-1}(f)\subseteq \operatorname{im}H_n(p)$. Therefore \begin{align*} \ker H_{n-1}(f)=\operatorname{im}H_n(p). \end{align*} [/step] [step:Identify the kernel of $H_{n-1}(i)$ with the image of $H_{n-1}(f)$] Let $[w]\in H_{n-1}(D)$, represented by a cycle $w\in D_{n-1}$ with $\partial^D_{n-1}w=0$. First suppose $[w]\in \operatorname{im}H_{n-1}(f)$. Then there is a cycle $x\in C_{n-1}$ such that $[w]=[f_{n-1}(x)]$. Since $\partial^C_{n-1}x=0$, \begin{align*} \partial^{\operatorname{Cone}}_n(0,x) &= \bigl(f_{n-1}(x),0\bigr)\\ &= i_{n-1}(f_{n-1}(x)). \end{align*} Thus $H_{n-1}(i)([f_{n-1}(x)])=0$, and therefore $[w]\in \ker H_{n-1}(i)$. Hence \begin{align*} \operatorname{im}H_{n-1}(f)\subseteq \ker H_{n-1}(i). \end{align*} Conversely, suppose $[w]\in \ker H_{n-1}(i)$. Then $[(w,0)]=0$ in $H_{n-1}(\operatorname{Cone}(f))$, so there exists $(y,x)\in \operatorname{Cone}(f)_n=D_n\oplus C_{n-1}$ such that \begin{align*} \partial^{\operatorname{Cone}}_n(y,x)=(w,0). \end{align*} Expanding the cone differential gives \begin{align*} \partial^D_n y+f_{n-1}(x)&=w,\\ -\partial^C_{n-1}x&=0. \end{align*} Thus $x$ is a cycle in $C_{n-1}$, and in $H_{n-1}(D)$ we have \begin{align*} [w]=[\partial^D_n y+f_{n-1}(x)]=[f_{n-1}(x)]. \end{align*} Therefore $[w]\in \operatorname{im}H_{n-1}(f)$, so \begin{align*} \ker H_{n-1}(i)\subseteq \operatorname{im}H_{n-1}(f). \end{align*} We conclude that \begin{align*} \ker H_{n-1}(i)=\operatorname{im}H_{n-1}(f). \end{align*} [/step] [step:Assemble the degreewise exactness identities into the long exact sequence] The previous three steps prove, for every $n\in\mathbb{Z}$, \begin{align*} \ker H_n(p)&=\operatorname{im}H_n(i),\\ \ker H_{n-1}(f)&=\operatorname{im}H_n(p),\\ \ker H_{n-1}(i)&=\operatorname{im}H_{n-1}(f). \end{align*} These are precisely the exactness conditions at the three consecutive terms \begin{align*} H_n(\operatorname{Cone}(f)),\qquad H_{n-1}(C),\qquad H_{n-1}(D). \end{align*} Since $n$ was arbitrary, the same argument applies in every degree. Hence the maps assemble into the long exact sequence \begin{align*} \cdots \longrightarrow H_n(D) \xrightarrow{H_n(i)} H_n(\operatorname{Cone}(f)) \xrightarrow{H_n(p)} H_{n-1}(C) \xrightarrow{H_{n-1}(f)} H_{n-1}(D) \xrightarrow{H_{n-1}(i)} H_{n-1}(\operatorname{Cone}(f)) \longrightarrow \cdots . \end{align*} This proves the theorem. [/step]