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

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You should add some connectedness assumptions, otherwise silly things can go wrong in degree 0. – Miha Habič Sep 2 '12 at 0:19
Even more worryingly, taking $X$ the wedge of infinitely many circles and $A$ and $B$ subcomplexes that leave out different finite numbers of circles, things go wrong. Maybe finite cell complexes work? – Miha Habič Sep 2 '12 at 0:38
It's true (for somewhat trivial reasons) if you modify your question slightly to ask about the category of topological pairs. 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! – Juan S Sep 2 '12 at 4:41

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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May be there are some conditions on the way $A$ and $B$ are included in $X$, under which the property is true, for example being "good pairs" As Hatcher calls them. For example when $A$ is contractible we can show using long exact sequ of pairs that $H_*(X,A)=\tilde H_*(X)$ which is no more then $H_*(X,*)$. – palio Sep 2 '12 at 6:11
@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. – Tyler Lawson Sep 2 '12 at 12:35