 Solution

Use integral singular chains. The inclusion $Y\subseteq X$ identifies $C_k(Y)$ with the subgroup of $C_k(X)$ generated by the singular simplices whose images lie in $Y$. It is a chain subcomplex. DefineThis boundary is well defined because $\partial C_k(Y)\subseteq C_{k-1}(Y)$, and its square is zero because ${\partial }^2=0$. Its homology is, by definition, $H_k(X,Y)$. A relative cycle has a representative $c$ with $\partial c\in C_{k-1}(Y)$; representatives of the same relative homology class differ by $\partial b+y$, where $b\in C_{k+1}(X)$ and $y\in C_k(Y)$.

Inclusion and quotient induce homomorphisms $i_{\ast }:H_k(Y)\to H_k(X)$ and $j_{\ast }:H_k(X)\to H_k(X,Y)$. Define the relative homology connecting homomorphism byHere $\partial c$ is a cycle in $Y$ since ${\partial }^{2}c=0$. If $c$ is changed to $c+\partial b+y$, its boundary changes by $\partial y$, a boundary in $Y$, proving independence of representatives. The formula is additive, so $\delta$ is a homomorphism.

The long exact sequence in relative homology isand ends inWe prove all three types of exactness directly, so no unproved abstract lemma about chain complexes is required.

At $H_k(X)$, $j_{\ast }{i}_{\ast }=0$ because a chain in $Y$ is zero in the relative chain group. Conversely, let an absolute cycle $c$ have $j_{\ast }[c]=0$. The definition of a relative boundary gives $c=\partial b+y$ with $y\in C_k(Y)$. Since $c$ is a cycle, $\partial y=0$. Thus $[c]=i_{\ast }[y]$, proving $\ker j_{\ast }=\mathrm{im}{i}_{\ast }$.

At $H_k(X,Y)$, $\delta j_{\ast }=0$ because an absolute cycle has zero boundary. Conversely, suppose $\delta [c]=0$. Then $\partial c=\partial y$ for some $y\in C_k(Y)$. Therefore $c-y$ is an absolute cycle and $j_{\ast }[c-y]=[c]$. Hence $\ker \delta =\mathrm{im}{j}_{\ast }$.

At $H_{k-1}(Y)$, $i_{\ast }\delta =0$ because $\partial c$ is a boundary in $X$. Conversely, if a cycle $y\in C_{k-1}(Y)$ becomes zero in $H_{k-1}(X)$, then $y=\partial c$ for some $c\in C_k(X)$. That $c$ is a relative cycle, and $\delta [c]=[y]$. Thus $\ker i_{\ast }=\mathrm{im}\delta$.

These arguments also prove the assertions at shifted indices. Finally, every zero-dimensional relative cycle lifts to a chain in $C_0(X)$, and every zero-chain is an absolute cycle; this proves surjectivity onto $H_0(X,Y)$. All maps, representative changes, compositions and exactness assertions have consequently been justified from the chain definitions.