Non-trivial $H_n(X,A)$ with trivial $H_n(X),H_n(A)$

Question

I am calculating the relative homology groups $H_n(S^3,S^1)$. From theorem 2.16 in Hatcher (relative homology groups fit in a long exact sequence) I get $H_2(S^3,S^1)\cong \mathbb{Z}$ since the relevant part of the long exact sequence is $$H_2(S^1)\to H_2(S^3) \to H_2(S^3,S^1) \to H_1(S^1) \to H_1(S^3)$$ Keeping in mind that $H_i(S^n)=\mathbb{Z}$ if $i=n,0$ and trivial otherwise we have $$0 \to 0 \to H_2(S^3,S^1) \to \mathbb{Z} \to 0$$ By exactness (I skip the details) $H_2(S^3,S^1) \cong H_1(S^1) \cong \mathbb{Z}$. My question is: the relative homology $H_n(X,A)$ somehow measures the difference between $H_n(X)$ and $H_n(A)$ but in this case, as $H_2(S^3)=0$, I might expect also $H_2(S^3,S^1)=0$. As it is not so I would like to understand what is measured by H_n(X,A) in the case where both previous homologies are $0$. Thanks

Answer 1

A very rough answer: absolute homology considers (homologically equivalent) cycles which are something like closed surfaces (perhaps self-intersecting). The relative homology considers something like surfaces which either have an empty boundary or it's boundary is contained in the smaller space.

In the example considered by OP, the membrane spanned by $\ S^1\ $ is a carrier of a relative 2-cycle.

Answer 2

Following your example, suppose only that $H_t(X)$ is zero. You can look at the exact sequence $$0 \to H_t(X,A) \stackrel{d}\to H_{t-1}(A)\stackrel{i}\to H_{t-1}(X) $$

Then the relative homology group $H_t(X,A)$ is the kernel of the inclusion $i$. Thus, it measures which $t$-cycles in $A$ are boundaries in $X$.