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I want to prove $$ \widetilde{ H_n}(X)\cong H_n(X, \ast)$$

The long exact sequence for pairs gives me $H_n(X) = \widetilde{H_n}(X) \cong H_n(X, \ast)$ for $n>0$

and $\widetilde{H_0}(X) \cong \widetilde{H_0}(X, \ast) $.

So I would like to show $\widetilde{H_0} (X, \ast) \cong H_0(X, \ast) $ to finish my proof.

Can anyone give me a hint on how to do this? Many thanks for your help.

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What is your definition of $\tilde{H}_n(X)$? –  Alexander Thumm Aug 22 '11 at 12:20
    
The homology groups that come from the extended chain complex where one inserts $\mathbb{Z}$ at the end of the sequence. (the definition in Hatcher) –  Matt N. Aug 22 '11 at 12:25
1  
Try to prove, that the relative chain complex $C(X,*)$ and the extended chain complex $\tilde C(X)$ are chain homotopic. –  Alexander Thumm Aug 22 '11 at 12:34
2  
Here is another way to look at the situation at hand: $i: (pt,\emptyset) \to (X,\emptyset)$ and $j: (X,\emptyset) \to (X,pt)$ give rise to a short exact sequence $E: \quad 0 \to C(pt) \to C(X), \to C(X,pt) \to 0$. The augmentation map $\epsilon : C_0(X) \to \mathbb Z \cong C_0(pt)$ then defines a left splitting of $E$. –  Alexander Thumm Aug 22 '11 at 13:07
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@Matt: Where did you get that sequence? Try using one of the sequences above - you should get a (split) five term short exact sequence –  Juan S Aug 23 '11 at 23:09
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1 Answer 1

up vote 1 down vote accepted

To elaborate on my third comment: The splitting defines an isomorphism $C(X) \cong C(X,*) \oplus C(*)$ with projections $\epsilon$ and $j$ and inclusion $i$. Using this identification we arrive at the following situation, where all the maps are the evident projections and inclusions:

$$\begin{array}{rcccl} \vdots & & \vdots & & \vdots \\ \tilde C_1(X) & \stackrel{p_{1}}{\cong} & C_1(X,*) & \stackrel{q_{1}}{\cong} & \tilde C_1(X) \\ \downarrow & & \downarrow & & \downarrow \\ \tilde C_0(X) \cong C_0(X,*) \oplus \mathbb Z & \stackrel{p_{0}}{\to} & C_0(X,*) & \stackrel{q_{0}}{\to} & C_0(X,*) \oplus \mathbb Z \cong \tilde C_0(X)\\ \downarrow & & \downarrow & & \downarrow \\ \mathbb Z & \stackrel{p_{-1}}{\to} & 0 & \stackrel{q_{-1}}{\to} & \mathbb Z \end{array}$$

It should now be easy to see that $h: \tilde C(X) \to \tilde C(X)^{+1}$, given by the inclusion $h_{-1}: \mathbb Z \to C_0(X,*) \oplus \mathbb Z$ and $h_k = 0$ otherwise, defines a chain homotopy of the horizontal map $q_* \circ p_*$ in the diagram above and the identity map on $\tilde C(X)$. On the other hand we have $p_* \circ q_* = id_{C(X,*)}$, which proves $C(X,*)$ and $\tilde C(X)$ to be chain homotopy equivalent.

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thanks! I understand that $0 \rightarrow C(\ast) \xrightarrow{i_\ast} C_0(X) \xrightarrow{j_\ast} C_0(X, \ast) \rightarrow 0$ splits because $i_\ast$ has an inverse (the constant map is a left inverse). From that it follows that $C_0(X) \cong C_0(X, \ast) \oplus C_0(\ast )$. How did you get $\tilde{C_0} (X) \cong C_0(X, \ast) \oplus \mathbb{Z} \cong C_0(X, \ast) \oplus C_0(\ast)$ from that? –  Matt N. Aug 25 '11 at 8:53
1  
No the left inverse is the augmentation map $\epsilon$ in dimension $0$ and constant map otherwise. Note that $\tilde C_k(X) = C_k(X)$ for all $k \geq 0$ and $\tilde C_{-1}(X) = \mathbb Z$ by definition. –  Alexander Thumm Aug 25 '11 at 9:24
    
But the augmentation map is at the end of the chain, like so: $0 \rightarrow C(\ast) \xrightarrow{i_\ast} C_0(X) \xrightarrow{j_\ast} C_0(X, \ast) \xrightarrow{\varepsilon} \mathbb{Z} \rightarrow 0$ so how can it be an inverse for $i_\ast$? –  Matt N. Aug 25 '11 at 9:36
    
$\begin{array}{ccccccccc} 0 & \to & C_0(\ast) & \stackrel{i_\ast}{\to} & C_0(X) & \to & C_0(X,\ast) & \to & 0 \\ & & \Arrowvert & & \Arrowvert & & & & \\ & & C_0(\ast)\cong \mathbb Z & \stackrel{\varepsilon}{\leftarrow} & C_0(X) & & & & \end{array}$ –  Alexander Thumm Aug 25 '11 at 10:28
    
While I don't know how to typeset this correctly... –  Alexander Thumm Aug 25 '11 at 10:28
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