# Definition of $\tilde{H}_n(X,A)$

My question arises from the following sentence of Hatcher's book p.118, in particular

I do understand that $$\tilde{H}_n(X,A)$$ is defined to be $$H_n(X,A)$$ if $$n \ne 0$$. There is a canonical way to define the reduced homology in order to have, given a long exact sequence of complex

$$0 \longrightarrow C_n(A) \longrightarrow C_n(X) \longrightarrow C_n(X,A) \longrightarrow 0$$

A natural long exact sequence of the pair $$(X,A)$$ for the reduced relative homology ? i.e

$$\cdots \longrightarrow \tilde{H}_n(A) \longrightarrow \tilde{H}_n(X) \longrightarrow H_n(X,A) \longrightarrow \tilde{H}_{n-1}(A) \longrightarrow \cdots$$

Related that doesn't solve the problem since the definition seems implicit.

• Does this not occur when we simply define the reduced homology as the homology of the augmented chain complex? Apr 12, 2021 at 8:52
• @memerson So the natural exact sequence for the relative "standard" homology of $(X,A)$, is the same since is transporded under this isomorphism ? Apr 12, 2021 at 8:58
• @memerson I think I don't get how to define the $0-$level $\tilde{H}_0(A) \longrightarrow \tilde{H}_0(X) \longrightarrow H_0(X,A) \longrightarrow 0$ since the other terms are equal to the non relative reduced version Apr 12, 2021 at 9:01
• There are two different steps in what is going on. First, $\tilde H_n(X,A) = H_n(X,A)$. Not just isomorphic. They are precisely equal as sets. They are identical in every possible way. Second, any short exact sequence of chain complexes gives rise to a natural long exact sequence in homology, so the short exact sequence of augmented chain complexes gives rise to a natural long exact sequence in reduced homology. But because $\tilde H(X,A) = H(X,A)$ we can put this in the exact sequence and preserve exactness and naturality. Apr 12, 2021 at 9:05
• I'm sure I could find a reference if you need one, but this happens because an augmented chain complex is still just a chain complex. So the same exact proof for the unaugmented version goes through. If the fact that we now have things in degree -1 is bothering you, we can simply reindex the chain complex, get our long exact sequence and then change the indices back Apr 12, 2021 at 9:11

The reduced homology groups of spaces $$X$$ and pairs $$(X,A)$$ are defined as the homology groups of the augmented chain complexes $$\tilde C_*(X)$$ and $$\tilde C_*(X,A)$$. For details see my answer to Suppose that $X$ is a topological space and $x_0\in X$. Prove that $\widetilde{H_n}(X)=H_n(X,x_0)$ for all $n\geq 0$. Note that $$X = \emptyset$$ plays a somewhat strange role in this context since formally $$\tilde H_{-1}(\emptyset) = \mathbb Z$$. Therefore one usually excludes the case $$X = \emptyset$$ when working with reduced homology. But note that formally everything works well even if we allow $$X = \emptyset$$.

In fact for any pair $$(X,A)$$ we obtain a short exact sequence of chain complexes $$0 \to \tilde C_*(A) \to \tilde C_*(X) \to \tilde C_*(X,A) \to 0 .$$ But each short exact sequence of chain complexes $$0 \to R \to S \to T \to 0$$ gives a long exact sequence of homology groups $$\dots \to H_n(R) \to H_n(S) \to H_n(T) \stackrel{\partial}{\to} H_{n-1}(R) \to \dots$$ This results in the natural long exact reduced homology sequence of the pair $$(X,A)$$.

From the definitions of $$\tilde C_*(X)$$ and $$\tilde C_*(X,A)$$ we see that

1. $$\tilde H_n(X) = H_n(X)$$ for $$n > 0$$

2. $$\tilde H_n(X,A) = H_n(X,A)$$ for all $$n$$.

3. $$H_0(X) \approx \tilde H_0(X) \oplus \mathbb Z$$

The long exact reduced homology sequence of a pair $$(X,A)$$ with $$A \ne \emptyset$$ is therefore identical with the long exact unreduced homology sequence except at the final terms: $$\dots \to H_1(A) \to H_1(X) \to H_1(X,A) \to \tilde H_0(A) \to \tilde H_0(X) \to H_0(X,A) \to 0$$

Also have a look at my answer to Clarification about reduced Homology.

• In 3. did you mean $n = 0$ ? Apr 12, 2021 at 9:02
• I'm sorry for the misunderstanding, my question was more addressed in understanding from where and the naturality of the long exact sequence for reduced homology you told me here comes from Apr 12, 2021 at 9:05
• @jacopoburelli Yes, in 3. I had a typo. The exactness of the long homology sequence (for any short exact sequence of chain complexes) is a purely algbraic result. See Hatcher p. 116. Apr 12, 2021 at 9:15