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?


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 $\tilde{H}_*(X,A)\cong H_*(X,A)$.

  • $\begingroup$ Thanks. Why do we need A to be non-empty? $\endgroup$
    – baby bunny
    Sep 12 '17 at 19:47
  • 1
    $\begingroup$ Well, Hatcher only defines reduced homology at all for non-empty spaces. But this isn't really necessary--the definitions still make perfect sense if you allow $A$ to be empty, and it's still true that $\tilde{H}_*(X,A)\cong H_*(X,A)$. (What is weird and presumably what Hatcher is trying to avoid is that $\tilde{H}_{-1}(\emptyset)\cong\mathbb{Z}$.) $\endgroup$ Sep 12 '17 at 19:53

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