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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.

Use integral singular homology. The singular chain group $C_n(B)$ is the free abelian group on continuous maps ${\Delta }^n\to B$, and its boundary operator isFor $i<j$, deleting vertex $j$ and then vertex $i$ has sign $(-1{)}^{i+j}$, while deleting $i$ and then the shifted vertex $j-1$ has sign $(-1{)}^{i+j-1}$. Both yield the same codimension-two face. Pairing these terms proves ${\partial }^2=0$. The inclusion $A\subset B$ identifies $C_n(A)$ with the subgroup spanned by those simplices having image in $A$, so it is injective and preserved by $\partial$.

The relative chain complex is $C_n(B,A)=C_n(B)/C_n(A)$ with induced boundary $\partial \bar{b}=\overline{\partial b}$. Thus there is a degreewise short exact sequence of chain complexesInclusion and quotient take homological cycles to homological cycles and boundaries to boundaries, hence define the group homomorphisms $i_{\ast }$ and $q_{\ast }$ on homology. We construct and prove exactness of the resulting sequence directly, rather than invoke an unproved algebraic lemma:ending in $H_0(A)\to H_0(B)\to H_0(B,A)\to 0$.

For the relative homology connecting homomorphism, lift a relative cycle $\bar{b}\in C_n(B,A)$ to $b\in C_n(B)$. Since $\partial \bar{b}=0$, the chain $\partial b$ belongs to $C_{n-1}(A)$; it is a homological cycle because ${\partial }^2=0$. DefineChanging the lift by $a\in C_n(A)$ changes the boundary by $\partial a$, hence does not change its homology class. More generally, if a different relative cycle represents the same relative homology class, lifts satisfy $b^{\prime }-b=\partial c+a$ for some $c\in C_{n+1}(B)$ and $a\in C_n(A)$. Then $\partial b^{\prime }-\partial b=\partial a$. This proves independence both of the lift and of the homology representative. Lifting a sum by the sum of lifts proves additivity, so $\delta$ is a well-defined group homomorphism.

At $H_n(A)$, an element in the image of $\delta$ is represented by $\partial b$ for some $b\in C_{n+1}(B)$, so its image in $H_n(B)$ is zero. Conversely, if a homological cycle in $A$ $a$ has $i_{\ast }[a]=0$, then $a=\partial b$ for a $B$-chain $b\in C_{n+1}(B)$. Its quotient $\bar{b}$ is a relative cycle and $\delta [\bar{b}]=[a]$. ThusFor completeness, the other terms are exact by the same elementary chain argument. At $H_n(B)$, clearly $q_{\ast }{i}_{\ast }=0$. If a homological cycle in $B$ $b$ has zero relative class, then $\bar{b}=\partial \bar{c}$ for a quotient chain $\bar{c}$. Lifting $c$ gives $a=b-\partial c\in C_n(A)$, and $\partial a=0$. Hence $[b]=i_{\ast }[a]$, proving $\ker q_{\ast }=\mathrm{im}{i}_{\ast }$. At $H_n(B,A)$, absolute homological cycles have zero connecting image. Conversely, if $\delta [\bar{b}]=0$, then $\partial b=\partial a$ for some $a\in C_n(A)$; the chain $b-a$ is an absolute homological cycle and maps to $[\bar{b}]$. Therefore $\ker \delta =\mathrm{im}{q}_{\ast }$. In degree zero every quotient chain lifts to an absolute homological cycle, because the differential to degree $-1$ is zero, so $q_{\ast }:H_0(B)\to H_0(B,A)$ is surjective. These proofs establish the whole long exact sequence in relative homology and all the algebraic facts used in its construction.