homology relative to homotopy equivalent subspaces let $A$ and $B$ are two subspaces of a topological space $X$. Is it true that if 
$A$ and $B$ are of the same homotopy type then the relative homology
$H_*(X,A;\mathbb Z)\cong H_*(X,B;\mathbb Z)$. More generally, if $A$ is a subspace of the top space $X$ and $B$ is a subspace of the top space $Y$
 such that $X$ and $Y$ have the same homotopy type and $A$ and $B$ have the same homotopy type then is it true that $H_*(X,A;\mathbb Z)\cong H_*(Y,B;\mathbb Z)$?
 A: The answer to the first question (and hence also to the second question) is no.
Let $X$ be the solid torus $S^1 \times D^2$, with $A = S^1 \times \{0\}$ and $B = \{1\} \times S^1$ (where we identify $S^1$ and $D^2$ with subspaces of the complex numbers).  Then $A$ and $B$ are homeomorphic, and hence of the same homotopy type.
The map $A \to X$ is the inclusion of the core of the solid torus, and you can retract the torus down to it.  This means it's a homotopy equivalence, and therefore the map $H_* A \to H_* X$ is an isomorphism.  The relative homology groups are zero.
The map $B \to X$ factors through an inclusion $B \to \{1\} \times D^2 \to X$, with the middle space contractible.  On $H_1$, this gives us a factorization $H_1 B \to 0 \to H_1 X$.  As a result, you can use the long exact sequence in homology to show $H_1(X,B) \cong H_2(X,B) \cong \mathbb{Z}$.
This illustrates an important aspect of homology theory.  It may initially seem like it's about associating abelian-group invariants to spaces, but it's even more fundamental that it tracks how functions relate them.
