If $A \subset X$, then there is an exact sequence which ends with ... 
If $A \subset X$, then there is an exact sequence $$\dots \rightarrow \tilde H_n(A) \rightarrow \tilde H_n(X) \rightarrow H_n(X,A) \rightarrow \tilde H_{n-1}(A) \rightarrow \dots $$
which ends
$$\dots \rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow H_0(X,A) \rightarrow 0$$
Hint:  $\tilde S_*(X) /\tilde S_*(A) = S_*(X) / S_*(A)$

I see that by the triple theorem, there is an exact sequence of this form for all $n \gt 0$ since $\tilde H_{n \gt 0}(X) = H_{n \gt 0}(X)$ and $\tilde H_0(X,A) = H_0(X,A)$, but for $n=0$ I don't see how to proceed.
Anyone have any ideas?
 A: Let $A \subset X$. We start with the reduced simplicial chain complexes:
$$ \cdots \xrightarrow{\partial} \tilde S_n(A) \xrightarrow{\partial} \tilde S_{n-1}(A) \xrightarrow{\partial} \cdots$$ 
$$\cdots \xrightarrow{\partial} \tilde S_n(X) \xrightarrow{\partial} \tilde S_{n-1}(X) \xrightarrow{\partial} \cdots $$
$$ \cdots \xrightarrow{\partial} \tilde S_n(X)/ \tilde S_n(A) \xrightarrow{\partial} \tilde S_{n-1}(X)/ \tilde S_{n-1}(A) \xrightarrow{\partial} \cdots $$
The maps $i: A \rightarrow X$, $j: X \rightarrow X/A$ give rise to a short exact sequence of chain complexes: 
$$
\begin{array}{ccccccccc}
   &  & 0 &  & 0 &  & 0 &  & \\
    & & \downarrow &  & \downarrow &  &  \downarrow & & \\
  \cdots & \xrightarrow{\partial} & \tilde{S}_{n+1}(A) & \xrightarrow{\partial} & \tilde{S}_n(A) & \xrightarrow{\partial} & \tilde{S}_{n-1}(A) & \xrightarrow{\partial} & \cdots \\
     & & \downarrow i_* &  & \downarrow i_* &  & \downarrow i_* & &  \\
  \cdots & \xrightarrow{\partial} & \tilde{S}_{n+1}(X) & \xrightarrow{\partial} & \tilde{S}_n(X) & \xrightarrow{\partial} & \tilde{S}_{n-1}(X) & \xrightarrow{\partial} & \cdots \\
   & & \downarrow j_* &  & \downarrow j_* &  & \downarrow j_* & &  \\
  \cdots & \xrightarrow{\partial} & \tilde{S}_{n+1}(X)/\tilde{S}_{n+1}(A) & \xrightarrow{\partial} & \tilde{S}_n(X)/\tilde{S}_n(A) & \xrightarrow{\partial} & \tilde{S}_{n-1}(X)/\tilde{S}_{n-1}(A) & \xrightarrow{\partial} & \cdots \\
    & & \downarrow &  & \downarrow &  & \downarrow & &  \\
   &  & 0 &  & 0 &  & 0 & &
\end{array}
$$
Utilizing the hint we replace $\tilde S_n(X)/\tilde S_n(A) \approx S_n(X)/S_n(A)$:
$$ \begin{array}{ccccccccc}
   &  & 0 &  & 0 &  & 0 &  & \\
    & & \downarrow &  & \downarrow &  &  \downarrow & & \\
  \cdots & \xrightarrow{\partial} & \tilde{S}_{n+1}(A) & \xrightarrow{\partial} & \tilde{S}_n(A) & \xrightarrow{\partial} & \tilde{S}_{n-1}(A) & \xrightarrow{\partial} & \cdots \\
     & & \downarrow i_* &  & \downarrow i_* &  & \downarrow i_* & &  \\
  \cdots & \xrightarrow{\partial} & \tilde{S}_{n+1}(X) & \xrightarrow{\partial} & \tilde{S}_n(X) & \xrightarrow{\partial} & \tilde{S}_{n-1}(X) & \xrightarrow{\partial} & \cdots \\
   & & \downarrow j_* &  & \downarrow j_* &  & \downarrow j_* & &  \\
  \cdots & \xrightarrow{\partial} & S_{n+1}(X)/S_{n+1}(A) & \xrightarrow{\partial} & S_n(X)/S_n(A) & \xrightarrow{\partial} & S_{n-1}(X)/S_{n-1}(A) & \xrightarrow{\partial} & \cdots \\
    & & \downarrow &  & \downarrow &  & \downarrow & &  \\
   &  & 0 &  & 0 &  & 0 & &
\end{array}$$
Which we use to obtain the long exact sequence:
$$\cdots \xrightarrow{\partial} \tilde H_n(A) \xrightarrow{i_*} \tilde H_n(X) \xrightarrow{j_*} H_n(X,A) \xrightarrow{\partial} \cdots $$
