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In Hatcher page 118, he says that

There is a completely analogous long exact sequence of reduced homology groups for a pair $(X;A)$ with $A\not = \emptyset$ ; This comes from applying the preceding algebraic machinery to the short exact sequence of chain complexes formed by the short exact sequences $0\rightarrow C_n(A)\rightarrow C_n(X)\rightarrow C_n(X;A)\rightarrow 0$ in nonnegative dimensions, augmented by the short exact sequence $0 \rightarrow\mathbb Z \stackrel{1}{\rightarrow} \mathbb Z \rightarrow 0 \rightarrow 0 $ in dimension $−1$. In particular, $\tilde H_n(X;A)$ is the same as $H_n(X;A)$ for all $n$, when $A\not = \emptyset$.

Could someone please explain how to derive the last conclusion which means that $\tilde H_0(X,A)\cong H_0(X,A)$. Indeed, for $n>0$ this isomorphism holds by construction of reduced homology. In reduced homology we have $$\cdots\rightarrow C_1(X,A)=C_1(X)/C_1(A)\stackrel{\partial_1}{\rightarrow}C_0(X,A)=C_0(X)/C_0(A)\stackrel{\epsilon}{\rightarrow}\mathbb Z \rightarrow 0$$ so $\tilde H_0(X,A)=\ker \epsilon /Im \partial_1$ while in non reduced homology we have $$\cdots\rightarrow C_1(X,A)=C_1(X)/C_1(A)\stackrel{\partial_1}{\rightarrow}C_0(X,A)=C_0(X)/C_0(A)\stackrel{\partial_0}{\rightarrow} 0$$ so $ H_0(X,A)=\ker \partial_0 /Im \partial_1$ why are these two groups isomorphic ?

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  • $\begingroup$ Actually they are not the same :) $\endgroup$ – Willard Zhan Jul 25 '15 at 10:06
  • $\begingroup$ @WillardZhan Hatcher is saying they are the same, I quoted the paragraph in my question. Could you please be more specific.. thanks! $\endgroup$ – palio Jul 25 '15 at 10:08
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The comment I made above misunderstood the problem, and I'm sorry for that...

To see they are the same, notice that $C_0(A)$ is not empty so the mapping $C_0(A)\stackrel{\epsilon}{\rightarrow}\mathbb{Z}$ is already surjective. Therefore $C_0(X,A)\stackrel{\epsilon}{\rightarrow}\mathbb{Z}$ is a zero mapping and its kernel is the same as $\ker \partial_0$.

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  • $\begingroup$ why is the mapping $C_0(A)\stackrel {\epsilon }{\rightarrow}\mathbb Z$ surjective ? and what is $\epsilon $ here ? $\endgroup$ – palio Jul 25 '15 at 11:07
  • $\begingroup$ I think you mean $C_0(A)\stackrel {\epsilon_A }{\rightarrow}\mathbb Z$ is surjective from the complex associated to $A$ and $C_0(X)\stackrel {\epsilon_X }{\rightarrow}\mathbb Z$ is surjective from the complex associated to $X$.. but $\epsilon_A$ is just the restriction homomorphism of $\epsilon_X$ hence the induced surjective homomorphism $\epsilon :C_0(X)/C_0(A)\stackrel{\epsilon}{\rightarrow} \mathbb Z$ must be zero. $\endgroup$ – palio Jul 25 '15 at 11:24

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