$H_n(X,A) \not \approx \tilde{H}_n(X/A)$ If the pair $(X,A)$ is good, then of course we have that $H_n(X,A) \approx \tilde{H}_n(X/A)$. For a counterexample to the converse, what is a nice example of a reasonably non-pathological bad pair $(X,A)$ such that $H_n(X,A) \not \approx  \tilde{H}_n(X/A)$ ? 
I would appreciate if $n = 1,2$ (so it can be visualized). 
 A: Let $X = S \cup A \subset \mathbb R^2$ be the topologist's sine curve, where
$$S = \{((x,\sin(1/x)) \mid 0 < x  \le 1 \} , A = \{0\} \times [-1,1] .$$
$X$ has two path components (these are $S$ and $A$) which are both contractible. In fact, $S$ is homeomorphic to $(0,1]$. Moreover, $X/A$ is homeomorphic to $[0,1]$, thus $\tilde H_n(X/A) = 0$ for all $n$. On the other hand consider the final part of the long exact sequence of $(X,A)$
$$H_0(A) \to H_0(X) \to H_0(X,A) \to 0 $$
We have $H_0(A) = \mathbb Z, H_0(X) = \mathbb Z \oplus \mathbb Z$. The map $H_0(A) \to H_0(X)$ embeds $\mathbb Z$ as a summand in $\mathbb Z \oplus \mathbb Z$. Thus we get a split short exact sequence
$$0  \to H_0(A) \to H_0(X) \to H_0(X,A) \to 0 $$
which implies that $H_0(X)  \approx H_0(A) \oplus H_0(X,A)$. Therefore $H_0(X,A) \ne 0$.
A: The suspension of $\{0,1,1/2,1/3, \dots \} $ has homology free on countably many generators because of the suspension isomorphism. The reduced suspension taking 0 as the basepoint is the Hawaiian earring. The homology of the Hawaiian earring is very difficult to compute because it is not semilocally simply connected. Suffice it to say that its homology is not free abelian.
Thus, since $H_n(X,A)=\bar{H}_n (X)$ if A is contractible (via the long exact sequence of a pair) the pair $X=S\{0,1,1/2,1/3, \dots \}$, $A= \{0\} \times I$ gives a counterexample because the first homology of the quotient is not the first homology of the original space.
A: Let $X=\Bbb S^2$ and $A=\Bbb S^2-\{\text{pt}\}$. Note that $A$ is not a sub-complex of $X$ as the open cell $\Bbb S^2-\{\text{pt}\}$ is contained in $A$ but the closure of $\Bbb S^2-\{\text{pt}\}$ is not contained in $A$. Note that $A\cong \Bbb R^2$. So, $A$ is contractible, and hence $H_2(A)=0=H_1(A)$. Next, $X/A$ is homeomorphic to the Sierpinski space. So, $X/A$ is contractible. Hence, $H_2(X/A)=0$.
Now, using the long exact sequence $\cdots \to H_2(A)\to H_2(X)\to H_2(X,A)\to H_1(A)\to \cdots$ we have $H_2(X,A)=H_2(X)=H_2(\Bbb S^2)=\Bbb Z$.

Let $X$ be the Sierpinski space, i.e. $X=\{x,y\}$ where the only open
sets are $X,\{x\},$ and $\emptyset.$ Now, consider the map $H:X\times
 [0,1]\to X$ defined by $$H:X\times
[0,1]\ni(z,t)\longmapsto\begin{cases}z &\text{ if }t=0,\\ x&\text{ if
 }t>0.\end{cases}$$ Now, $H^{-1}\big(\{x\}\big)=\big(X\times
(0,1]\big)\cup \big(\{x\}\times 0\big)=\big(\{x\}\times [0,1]\big)\cup \big(\{y\}\times (0,1]\big)$, and its complement in $X\times [0,1]$ is
$\{y\}\times \{0\}$. So, $H$ is continuous. Hence, $X$ is
contractible.

