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 ?

  • $\begingroup$ Actually they are not the same :) $\endgroup$
    – Wei Zhan
    Jul 25, 2015 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, 2015 at 10:08

2 Answers 2


I know this is a very late answer, but I think there is a basic confusion here, which I don't want to let stand. The other answer is very confusing (at least, to me) and since the result almost trivial once one realizes that the OP's reduced relative sequence is wrong, I think it is worthwhile to write down the simple proof.

Indeed, since $A\subseteq X$, and since the reduced chains for $A$ and $X$ are

$\cdots \longrightarrow C_1(A)\longrightarrow C_0(A)\longrightarrow\mathbb{Z}\longrightarrow0\quad$ and $\quad \cdots \longrightarrow C_1(X)\longrightarrow C_0(X)\longrightarrow\mathbb{Z}\longrightarrow0$

by definition the relative chain is given by

$\cdots \longrightarrow C_1(X)/C_1(A)\longrightarrow C_0(X)/C_0(A)\longrightarrow\mathbb{Z}/\mathbb{Z}\longrightarrow0=$

$\cdots \longrightarrow C_1(X)/C_1(A)\longrightarrow C_0(X)/C_0(A)\longrightarrow 0\longrightarrow0$.

But this last chain is exactly the same as the one induced by the unreduced chains

$\cdots \longrightarrow C_1(A)\longrightarrow C_0(A)\longrightarrow0\quad$ and $\quad \cdots \longrightarrow C_1(X)\longrightarrow C_0(X)\longrightarrow0$

so $H(X,A)$ and $\widetilde H(X,A)$ are the same.

  • $\begingroup$ Thanks,excellent explanation. As Wei Zhan mentioned for the relative chain to satisfy $C_0(X)/C_0(A) \rightarrow 0$, it requires $A$ not being empty right? And the long exact sequence for reduced homology group is $\cdots \stackrel{\partial}{\rightarrow} \widetilde{H}_{0}(A) \stackrel{i_*}{\rightarrow} \widetilde{H}_{0}(X) \stackrel{j_*}{\rightarrow} \widetilde{H}_{0}(X,A)\stackrel{\partial}{\rightarrow} \Bbb{Z} {\rightarrow} \Bbb{Z} {\rightarrow} 0$? $\endgroup$
    – onriv
    Feb 7 at 15:37
  • $\begingroup$ It should be $\ldots \stackrel{\partial}{\rightarrow} \widetilde{H}_{0}(A) \stackrel{i_{*}}{\rightarrow} \widetilde{H}_{0}(X) \stackrel{j_{*}}{\rightarrow} \widetilde{H}_{0}(X, A) \stackrel{\partial}{\rightarrow} \widetilde{H}_{-1}(A) \rightarrow \widetilde{H}_{-1}(X) \rightarrow 0.$ $\endgroup$
    – Sushant
    Jun 24 at 21:01

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 $\ker \partial_0$.

  • $\begingroup$ why is the mapping $C_0(A)\stackrel {\epsilon }{\rightarrow}\mathbb Z$ surjective ? and what is $\epsilon $ here ? $\endgroup$
    – palio
    Jul 25, 2015 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, 2015 at 11:24

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