= 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. Define
$$
C_k(X,Y)=C_k(X)/C_k(Y),\qquad
\bar\partial(c+C_k(Y))=\partial c+C_{k-1}(Y).
$$
This 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 (mathematics)> 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 $i_*:H_k(Y)\to H_k(X)$ and $j_*:H_k(X)\to H_k(X,Y)$. Define the <relative homology connecting homomorphism> by
$$
\delta:H_k(X,Y)\longrightarrow H_{k-1}(Y),
\qquad \delta[c]=[\partial c].
$$
Here $\partial c$ is a cycle in $Y$ since $\partial^2c=0$. 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> is
$$
\boxed{\cdots\longrightarrow H_k(Y)
\xrightarrow{i_*}H_k(X)
\xrightarrow{j_*}H_k(X,Y)
\xrightarrow{\delta}H_{k-1}(Y)
\xrightarrow{i_*}H_{k-1}(X)\longrightarrow\cdots}
$$
and ends in
$$
H_0(Y)\xrightarrow{i_*}H_0(X)
\xrightarrow{j_*}H_0(X,Y)\longrightarrow0.
$$
We prove all three types of exactness directly, so no unproved abstract lemma about <chain complexes> is required.
At $H_k(X)$, $j_*i_*=0$ because a chain in $Y$ is zero in the relative chain <group>. Conversely, let an absolute cycle $c$ have $j_*[c]=0$. 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 $[c]=i_*[y]$, proving $\ker j_*=\operatorname{im}i_*$.
At $H_k(X,Y)$, $\delta j_*=0$ 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 $j_*[c-y]=[c]$. Hence $\ker\delta=\operatorname{im}j_*$.
At $H_{k-1}(Y)$, $i_*\delta=0$ 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 $\ker i_*=\operatorname{im}\delta$.
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.
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