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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 $H_*(X,A;\mathbb Z)\cong H_*(Y,B;\mathbb Z)$?

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    It's true (for somewhat trivial reasons) if you modify your question slightly to ask about the category of [topological pairs](http://en.wikipedia.org/wiki/Topological_pair). I say trivial reasons, for if you think of homology as something satisfying the Eilenberg-Steenrod axioms, then, by definition, homotopic maps induce the same maps in homology. Anyway - you may already know this, but +1 for the nice question!2012-09-02

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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.

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    @palio, that's partially because the homology of a point is so simple. Both of the inclusions that I listed are "good" in Hatcher's sense. The easiest situation that I know where what you want is true is if there is a map $A \to B$, which is a homotopy equivalence, such that the inclusion $A \subset X$ and the composite $A \to B \subset X$ are homotopic.2012-09-02