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In the section Axioms for homology from Hatcher's Algebraic Topology (page 161) he says:

Note that $\tilde{h}_n(x_0) = 0$ for all $n$, as can be seen by looking at the long exact sequence of reduced homology groups of the pair $(x_0,x_0)$.

I guess that he means the exact sequence of the second axiom, which gives the boundary map $\partial :\tilde{h}_n(X/A)\to\tilde{h}_{n-1}(A)$.

From this it is clear that $\tilde{h}_n(x_0)=\tilde{h}_{n-1}(x_0)$ for all $n$. But Hatcher is not assuming the dimension axiom, so I cannot say that $\tilde{h}_n(x_0)=0$ for $n\neq 0$, so how could I conclude that $\tilde{h}_n(x_0) = 0$ for all $n$?

I've thought that this could be shown using the first axiom if I could show that constant maps induce trivial maps in homology, but I don't see how to deduce this fact from the axioms because the proof I know for singular, cellular or simplicial homology uses that the homology of a point is trivial.

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    $\begingroup$ I'm not sure how you concluded that $\tilde{h}_n(x_0)=\tilde{h}_{n-1}(x_0)$. There is a boundary map between them, but why would that map be an isomorphism? (In fact it is since both groups are trivial, but I don't know what logic you're using.) $\endgroup$ – Eric Wofsey Dec 22 '18 at 18:47
  • $\begingroup$ @EricWofsey in this case, $X=A=x_0$, so $X/A=A$, and the boundary map is an isomorphism. $\endgroup$ – Javi Dec 22 '18 at 18:49
  • $\begingroup$ But...why is it an isomorphism? $\endgroup$ – Eric Wofsey Dec 22 '18 at 18:50
  • $\begingroup$ @EricWofsey I was going to write my argument but I found the mistake, thanks. $\endgroup$ – Javi Dec 22 '18 at 18:54
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We have an exact sequence $$\tilde{h}_n(A)\stackrel{i_*}\to\tilde{h}_n(X)\stackrel{q_*}\to\tilde{h}_n(X/A)$$ where the first map is induced by the inclusion $i:A\to X$ and the second map is induced by the quotient map $q:X\to X/A$. But if $X$ and $A$ are both just a point, then $i$ and $q$ are both homeomorphisms, so $i_*$ and $q_*$ are both isomorphisms. But the image of $i_*$ is the kernel of $q_*$, so this implies $q_*=0$, and so $\tilde{h}_n(X)=0$ since $q_*$ is an isomorphism.

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Hatcher's definition of a reduced homology theory is limited to CW-pairs. A more general definition for general pairs is this: Omit axiom 3 and replace the long exact sequence in axiom 2 by $$\dots \stackrel{\partial}{\rightarrow} \tilde{h}_n(A) \stackrel{i_*}{\rightarrow} \tilde{h}_n(X) \stackrel{j_*}{\rightarrow} \tilde{h}_n(X \cup CA) \stackrel{\partial}{\rightarrow} \dots$$ Here, $CA$ is the unreduced cone on $A$ and $X \cup CA$ is the adjunction space in which the equivalence class $[a,0] \in CA$ in the base of $CA$ is identified with $a \in A \subset X$. For CW-pairs this is equivalent to Hatcher's long exact sequence. In fact, if $(X,A)$ is CW-pair, then $(X \cup CA, CA)$ is also a CW-pair, in a particular $CA \hookrightarrow X \cup CA$ is cofibration. Since $CA$ is contractible, the quotient map $p : X \cup CA \to (X \cup CA)/CA = X/A$ is a homotopy equivalence.

Now consider $(X,A) = (x_0,x_0)$. Then $X \cup CA$ is homeomorphic to the unit interval whence $j_*$ is an isomorphism. Since $i_*$ trivially is an isomorphism, we conclude that $\tilde{h}_n(x_0) = 0$ for all $n$.

Added:

Hatcher's approach to axiomatic homology is not standard. In fact, he gives an axiomatic definition of a reduced single space homology theory for unbased spaces. This means that no relative groups $\tilde{h}_n(X,A)$ are used.

Such theories for unbased spaces are considered only rarely in the literature (but is a common to work with reduced single space homology theory for based spaces, although their definition is different from Hatchers's).

If you consider singular homology defined for all pairs of toplogical spaces, then it is well known that reduced singular homology groups can be defined via the augmented chain complex.

A more general approach is this: If we have a generalized homology theory $(H_n,\partial)$ satisfying the Eilenberg-Steenrod axioms with the possible exception of the dimension axiom, we can define reduced groups by $$\tilde{H}_n(X) = \ker (p_* : H_n(X) \to H_n(*)) \subset H_n(X)$$ where $p : X \to *$ is the unique map to a one-point space $*$. It is then an easy exercise to show that we get a long exact sequence $$\dots \stackrel{\partial}{\rightarrow} \tilde{H}_n(A) \stackrel{i_*}{\rightarrow} \tilde{H}_n(X) \stackrel{j_*}{\rightarrow} H_n(X,A) \stackrel{\partial}{\rightarrow} \tilde{H}_{n-1} (A)\stackrel{i_*}{\rightarrow} \dots $$ The excision axiom shows that $H_n(X,A) \approx H_n(X \cup CA,CA)$ and the latter is easily seen to be isomorphic to $\tilde{H}_n(X \cup CA)$. This yields the exact sequence from above. Note that it does not agree with Hatcher's exact sequence, except if restrict to CW-pairs.

In the case of reduced singular homology the definition via the augmented chain complex and our general definition agree and we have $\tilde{H}_n(X) = H_n(X)$ for $n > 0$. But reduced singular homology does not satisfy Hatchers's axiom 2 (long exact sequence with quotients $X/A$). Consider for example the pair $(T,J)$ where $T = \{ x,\sin(1/x)) \mid x \in (0,1] \} \cup \{ 0 \} \times [-1,1]$ is the closed topologist's sine curve and $J = \{ 0 \} \times [-1,1]$. The space $T$ has two path components which are contractible and $T/J$ is homeomorphic to the interval $[0,1]$, i.e. contractible. Hence $\tilde{H}_0(J) = 0, \tilde{H}_0(T) = \mathbb {Z}, \tilde{H}_0(T/J) = 0$ which shows that $\tilde{H}_0(J) \stackrel{i_*}{\rightarrow} \tilde{H}_0(T) \stackrel{p_*}{\rightarrow} H_0(T/J)$ cannot be exact.

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  • $\begingroup$ Thanks for your answer! $\endgroup$ – Javi Dec 23 '18 at 11:57
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    $\begingroup$ @Javi I added some material. $\endgroup$ – Paul Frost Dec 23 '18 at 15:06

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