The following sentence appears on page 118 of Hatcher's Algebraic Topology:
In particular
this means that $\tilde H_n (X, A)$ is the same as $H_n (X, A)$ for all n , when $A\neq \emptyset$ .
However I can't find the definition of $\tilde H_n (X, A)$. Is it defined to be the homology group associated to the chain complex
$$\cdots\rightarrow C_k(X,A)\rightarrow C_{k-1}(X,A)\rightarrow\cdots\rightarrow C_0(X,A)\rightarrow \mathbb{Z}?$$
If so why should they be equal?
Best Answer
The definition is in the sentence right before the statement. Specifically, we can consider the augmented chain complexes $$\dots \to C_2(A)\to C_1(A)\to C_0(A)\to\mathbb{Z}\to 0$$ and $$\dots \to C_2(X)\to C_1(X)\to C_0(X)\to \mathbb{Z}\to 0$$ used to compute $\tilde{H}_*(A)$ and $\tilde{H}_*(X)$. We can naturally consider the first chain complex as a subcomplex of the second chain complex, and form the quotient complex. But this quotient complex will be exactly the same as the ordinary relative chain complex $C_*(X,A)$, since the augmentation "cancels out" (when we take the quotient of $\mathbb{Z}$ by $\mathbb{Z}$, we get $0$). So the chain complex which computes $\tilde{H}_*(X,A)$ is just $$\dots\to C_2(X,A)\to C_1(X,A)\to C_0(X,A)\to 0 \to 0$$ and so $\tilde{H}_*(X,A)$ is the same thing as $H_*(X,A)$.
The point of this construction is that these three chain complexes form a short exact sequence of chain complexes. As a result, we get a long exact sequence relating their homologies: that is, relating $\tilde{H}_*(A)$, $\tilde{H}_*(X)$, and $\tilde{H}_*(X,A)\cong H_*(X,A)$.