If $A/B$ is homeomorphic to $C/D$, where $B\subset A$ and $D\subset C$, then is $H_i(A,B)=H_i(C,D)$? On pg 125, in his book "Algebraic Topology" Hatcher makes a claim that is analogous to the following:
If $A/B$ is homeomorphic to $C/D$, where $B\subset A$ and $D\subset C$, then $H_i(A,B)=H_i(C,D)$. 
Is this true? And if it is, why is this true? I know that $H_i(A/B)\neq H_i(A,B)$ (or at least this is what I understand). 
 A: It is false in general, for much the same reason that "$H_k(X, Y) \cong \tilde{H}_k(X/Y)$" isn't always true.
Here's a very non-pathological counter-example: for $n> 0$ let $X = \mathbb{R}^n$ and $Y = \mathbb{R}^n \setminus \{0\}$. Then $X/Y$ is homeomorphic to the Sierpinski space, which is the unique topology on the two point set $\{o, x\}$ such that $\{o\}$ is open and $\{x\}$ is not, and this space is in fact contractible. On the other hand, using the fact that $\mathbb{R}^n \setminus \{0\} \simeq S^{n-1}$ you can compute via the long exact sequence of the pair that $H_k(X,Y)$ is $\mathbb{Z}$ if $k=n$ and $0$ otherwise. (In particular $(X,Y)$ satisfies $H_n(X,Y) \neq \tilde{H}_n(X/Y)$.)
Now let $n_1 \neq n_2 > 0$, let $(A,B) = (\mathbb{R}^{n_1}, \mathbb{R}^{n_1} \setminus\{0\})$, and let $(C,D) = (\mathbb{R}^{n_2}, \mathbb{R}^{n_2} \setminus\{0\})$. Then both quotients $A/B$ and $C/D$ are homeomorphic to the Sierpinski space, but the pairs have different relative homologies.
Like Elliot G points out in their comment, you need a notion of "good pair". If $(A,B)$ and $(C,D)$ are both good in the sense of Hatcher then the statement you want is true, since in this case we would have
$$H_*(A,B) \stackrel{good pair}{\cong} \tilde{H}_*(A/B) \stackrel{assumption}{\cong} \tilde{H}_*(C/D) \stackrel{good pair}{\cong} H_*(C,D) $$
