I read in Hatcher page 124 that

relative homology can be expressed as reduced absolute homology in the case of good pairs $(X,A)$, but in fact there is a way of doing this for arbitrary pairs.

His argument goes as follows: $\displaystyle \tilde{H_n}(X/A)= \tilde{H_n}(X ∪ CA) = H_n(X ∪ CA,CA) = H_n(X ∪ CA − \{p\},CA − \{p\}) = H_n(X,A)$

Where $p$ is the apex of $CA$ the cone on $A$.

So I understand that the relation $\tilde{H_n}(X/A)=H_n(X,A)$ holds without any condition on the pair $(X,A)$ where $X$ is a topological space and $A$ is any non empty subset of $X$. Is my understanding correct ? thank you for your help!


No, this is a misunderstanding.

Hatcher says that you can identify the relative homology group $H_n(X,A)$ of a good pair $(X,A)$ with the reduced homology group $\tilde{H}_n(X/A)$ of the space $X/A$. Good pairs are defined as those satisfying the hypothesis of Theorem 2.13. For example, CW-pairs are good.

This means that the homology groups of good pairs can be expressed as the (reduced) homology groups of suitable single spaces. This is possible for arbitrary pairs, though in general not in the form $\tilde{H}_n(X/A)$, but in the form $\tilde{H}_n(X \cup CA)$ with the single space $X \cup CA$. There is an obvious quotient map $X \cup CA \to X/A$ which is a homotopy equivalence for good pairs (but not for general pairs).

  • $\begingroup$ Thank you @Paul Frost. I see your point.. what about the identification $X/A=(X\cup CA)/CA$ is this true for all pairs $(X,A)$ or only for good pairs ? $\endgroup$
    – palio
    Jun 20 '18 at 15:27
  • 1
    $\begingroup$ That is true for all pairs, but unfortunately the quotient map $p : X \cup CA \to (X \cup CA)/CA = X/A$ is in general not a homotopy equivalence. One can show that a quotient map $q : Y \to Y/B$ is a homotopy equivalence provided $B$ is contractible and the inclusion map $i : B \to Y$ is a cofibration. $\endgroup$
    – Paul Frost
    Jun 20 '18 at 15:34
  • $\begingroup$ To see that $X/A = (X \cup CA)/CA$, note that there is a natural embedding $j : X \to X \cup CA$.Let $r : X \cup CA \to (X \cup CA)/CA$ denote the quotient map. Then $r \circ j$ maps $A$ to the point obtained by collapsing $CA$, therefore it induces a continuous map $j' : X/A \to (X \cup CA)/CA$. Clearly $j'$ is a bijection and it is easy to show that it is an open map (i.e. a homeomorphism). $\endgroup$
    – Paul Frost
    Jun 20 '18 at 15:49

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