If $A$ is a weak deformation retraction of $X$ then the relative homology $H_n(X,A)$ is trivial for all $n$

Question

I want to prove that if $A$ is a weak deformation retraction of $X$, then the relative homology $H_n(X,A)$ is trivial for all $n$.

I would like to prove this by induction. Studying the LES, I get:

$\dots \to H_1(X,A) \to H_0(A) \to H_0(X) \to H_0(X,A) \to 0$.

I thought I could argue that $H_0(A) = H_0(X)$ because $A$ and $X$ are homotopic, but I'm not sure. Anyway, not even using that I could show what I wanted. I'm a little beginner in the subject. Could someone provide me with some insight? Thanks!

Answer 1

Consider the inclusion map $i:A\hookrightarrow X$, which is an homotopy equivalence by hypothesis.

We have the following short exact sequence of chain complex (of abelian groups): $$ 0 \to C_*(A) \hookrightarrow C_*(X)\to C(X)_*/C_*(A) \to 0 $$ By a basic result from homological algebra, the above short exact sequence induces the following long exact sequence at the level of the homology: $$ \ldots\to H_{*+1}(X,A) \to H_*(A) \to H_*(X)\to H_*(X,A)\to H_{*-1}(A)\to\ldots $$

Now, the arrow $H_*(A) \to H_*(X)$ was induced by the homotopy equivalence $i:A\hookrightarrow X$ and so, by a basic result from algebraic topology, it is an isomorphism. By the exactness of the sequence above, $H_*(X,A)=0$.