In example 2.18 page 118 of Hatcher's Algebraic topology we read :

Applying the long exact sequence of reduced homology groups to a pair $(X,x_0)$ with $x_0\in X$ yields isomorphisms $H_n(X,x_0)=\tilde H_n (X)$ for all $n$ since $\tilde H_n (x_0)=0$ for all $n$.

Actually the long exact sequence gives that $\tilde H_n(X,x_0)=\tilde H_n (X)$ for all $n$, i think it remains to show that $\tilde H_n(X,x_0)=H_n(X,x_0)$ for all $n$ which is equivalent to show that $\tilde H_0(X,x_0)=H_0(X,x_0)$ but how to do it? and is it true more generally that $\tilde H_0(X,A)=H_0(X,A)$ for all non empty $A$?


1 Answer 1


Right before Example 2.17 on the same page, Hatcher says

In particular this means that $\tilde{H}_n(X, A)$ is the same as $H_n(X, A)$ for all $n$, when $A \neq \emptyset$.

  • $\begingroup$ thank you for pointing that out!! why is this true? what is the idea behind that? $\endgroup$
    – palio
    Commented Sep 26, 2012 at 20:00
  • 3
    $\begingroup$ The idea is explained in that paragraph of the text: essentially, relative chain complexes are already augmented. $\endgroup$ Commented Sep 26, 2012 at 20:11

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