$H_{n}(X - W , A-W) \cong H_{n}(X,A)$

Question

Let $(X,A)$ be a pair with $A \subset X$ ( topology space) and let $W$ be a subset of $A$ such that $\overline{W}$ is contained in interior $A$ . Prove that the inclusion map induces an isomorphism of relative homology groups : $$H_{n}(X - W , A - W ) \cong H_{n}(X,A) \forall n \in N$$

Answer 1

The key result you need is this:

If $U$ and $V$ are subspaces of $X$ such that the interiors of $U$ and $V$ cover $X$, then the inclusion map $V \to X$ induces an isomorphism $H_n(V, U \cap V) = H_n(X, U)$.

Your claim follows immediately by taking $U = A$ and $V = X - W$.

I strongly recommend reading the proof of this result in Hatcher on pages 119-124. Admittedly, Hatcher's proof looks a bit technical at first sight, but the idea is really very simple!

The main idea in Hatcher's proof is the following: Every singular simplex in $X$ can be subdivided into a collection of smaller simplices that are small enough that each of them either fits inside $U$ or fits inside $V$.

For example, suppose $X = [0,1], U = [0,\frac 2 3), V = (\frac 1 3, 1]$. Suppose that you are given the $1$-simplex that is mapped homeomorphically onto the interval $[\frac 1 4, \frac 3 4]$. At the moment, this simplex does not fit into either $U$ or $V$. However, you can split up this simplex into two smaller simplices, mapped homeomorphically onto $[\frac 1 4, \frac 1 2]$ and $[\frac 1 2, \frac 3 4]$ respectively. Notice that the $[\frac 1 4, \frac 1 2]$ simplex is contained inside $U$ and the $[\frac 1 2, \frac 3 4]$ simplex is contained inside $V$. This illustrates the claim above.

Thus we have a way of turning singular chains in $C_n(X)$ into singular chains in $C_n(U+V)$, where $C_n(U + V)$ denotes the free abelian group generated by "little" singular simplices that are small enough to fit inside either $U$ or $V$.

Of course, the new singular chain in $C_n(U + V)$ that you get after this subdivision process is different from the original singular chain in $C_n(X)$ that you started with. Nonetheless, we have:

For the purpose of computing homology, the new subdivided chain in $C_n(U + V)$ is equivalent to the original chain in $C_n(X)$.

(Intuitively this makes sense. The original chain and the subdivided chain "occupy the same space".)

So how does this help? Well, by definition, the $H_n(X, U)$'s are the homology groups for the complex $$ \left( \frac{C_n(X)}{C_n(U)}, \partial_n \right).$$ But, as we said above, it is possible to replace $C_n(X)$ with $C_n(U + V)$ for the purposes of computing homology. So we can equally compute the $H_n(X, U)$'s by taking the homology of the complex $$ \left( \frac{C_n(U + V)}{C_n(U)}, \partial_n \right).$$ It shouldn't be hard to see that $C_n(U + V) / C_n(U)$ is the same thing as $C_n(V) / C_n(U \cap V).$ Thus the $H_n(X, U)$'s are the homology groups of the complex $$ \left( \frac{C_n( V)}{C_n(U \cap V)}, \partial_n \right).$$ But the $n$th homology group of this complex is $H_n(V, U \cap V)$, by definition! Hence, $$ H_n(V, U \cap V) = H_n(X, U).$$

To make this argument rigorous, you need to show that (i) this subdivision process is always possible and (ii) the subdivided chain really is equivalent to the original chain for the purposes of computing homology. These technical issues are dealt with in Hatcher. The subdivision process is called "barycentric subdivision", and the claim that the original and subdivided chains are equivalent in homology is made precise using an algebraic tool called a "chain homotopy". Good luck!