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

enter image description here

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.

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  • $\begingroup$ Does this not occur when we simply define the reduced homology as the homology of the augmented chain complex? $\endgroup$
    – memerson
    Apr 12, 2021 at 8:52
  • $\begingroup$ @memerson So the natural exact sequence for the relative "standard" homology of $(X,A)$, is the same since is transporded under this isomorphism ? $\endgroup$ Apr 12, 2021 at 8:58
  • $\begingroup$ @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 $\endgroup$ Apr 12, 2021 at 9:01
  • $\begingroup$ 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. $\endgroup$
    – memerson
    Apr 12, 2021 at 9:05
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    $\begingroup$ 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 $\endgroup$
    – memerson
    Apr 12, 2021 at 9:11

1 Answer 1

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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.

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  • $\begingroup$ In 3. did you mean $n = 0$ ? $\endgroup$ Apr 12, 2021 at 9:02
  • $\begingroup$ 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 $\endgroup$ Apr 12, 2021 at 9:05
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    $\begingroup$ @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. $\endgroup$
    – Paul Frost
    Apr 12, 2021 at 9:15

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