Suppose Inclusion is a homotopy equivalence

Question

Suppose inclusion $j:A->X$ is a homotopy equivalence. Show $H_p(X,A)=0$.

So I'm new to this algebraic topology stuff, but I think the idea here is that since $j$ is a homotopy equivalence, that the group $H_p(X,A)$ would be zero because all the cycles of $X$ would be modded out by the cycles of $A$... I don't know how to show this though.

Answer 1

If $j:A\to X$ is a homotopy equivalence, then it induces isomorphisms in homology: $j_*:H_p(A)\to H_p(X)$ is an isomorphism for all $p$. Considering now the long exact sequence of the pair $(X,A)$ gives $H_p(X,A)=0$ for all $p$.