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Let $X$ be a non-empty topological space and consider $x \in X$.

Consider the following sequence of homology groups:

$$0 \to \tilde{H}_0(X) \to H_0(X) \to H_0(\{x\}) \to 0$$

Here the map $i: \tilde{H}_0(X) \to H_0(X)$ is the inclusion map and the map $c_*: H_0(X) \to H_0(\{x\})\cong \mathbb{Z}$ is the map induced by the map $c: X \to \{x\}: y \mapsto x$.

In particular, I can see that the left map is injective and the right map is surjective, so I only need to understand why the sequence is exact at $H_0(X)$?

So basically, my question is: why is $\tilde{H}_0(X) = \ker (c_*)$?

Shaun
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  • what is $\tilde{H_0}(X)$? – Just dropped in Jun 28 '20 at 19:20
  • The reduced homology group. –  Jun 28 '20 at 19:24
  • $\widetilde H_nX=\ker(H_nX\rightarrow H_n\ast)$ is really the most sensible definition of the reduced homology. – Tyrone Jun 28 '20 at 19:39
  • @Tyrone I follow the definition in Hatcher: $\tilde{H}_0(X) = \ker\omega/\operatorname{Im}(\partial_1)$ where $\omega: C_0(X) \to \mathbb{Z}$ is the morphism that sends $\sigma$ to $1$. –  Jun 28 '20 at 20:26
  • Do you know that $\tilde{H}_0(X) \cong H_0(X,x_0)$? I'm guessing not because then it's immediate. – Osama Ghani Jun 28 '20 at 20:31
  • I know that but I prefer not to use it as relative homology groups are introduced much later in the text. –  Jun 28 '20 at 20:32
  • If you want to use that definition, then consider this: think of the augmentation as a map of (unaugmented) chain complexes. It is the chain map induced by $X\rightarrow\ast$. If $X$ is nonempty, then there is a (non-canonical) embedding in the other direction and you can study the long exact sequence of the pair $(X,\ast)$. Compare this with the previous map. – Tyrone Jun 28 '20 at 20:54

1 Answers1

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You probably know that Eilenberg and Steenrod introduced axioms to define the concept of a homology theory on an abstract level. Singular homology is one example of such a theory, but there are many other.

Given such a homology theory, a general method to define reduced homology groups is $$(D) \quad \tilde H_n(X) = \ker (c_* : H_n(X) \to H_n(*))$$ where $c : X\to *$ denotes the unique map to the one-point space $*$. See Tyrone's comment.

However, the standard "textbook approach" for singular homology is to define reduced homology groups $\tilde H_n(X)$ as the homology groups of the augmented chain complex. Doing so, equation $(D)$ becomes a theorem which requires a proof.

Working with the augmented chain complex yields $\tilde H_n(X) = H_n(X)$ for $n > 0$ (both are the same quotient groups), thus trivially $(D)$ is satisfied for $n > 0$ since $H_n(*) = 0$. In dimension $0$ let us observe that $\tilde H_0(X)$ is (in contrast to the kernel definition) not a genuine subgroup of $H_0(X)$, but there is a canonical group monomorphism $\iota : \tilde H_0(X) \to H_0(X)$: The first group is $\ker \epsilon / \text{im} \partial_1$, the second is $C_0(X) / \text{im} \partial_1$ and $\iota$ is induced by the inclusion $\mu : \ker \epsilon \hookrightarrow C_0(X)$. If we understand this point, we may laxly write $\tilde H_0(X) \subset H_0(X)$. However, the precise statement is that $\text{im} \iota = \ker c_*$.

Consider the induced map $c_\# : C_0(X) \to C_0(*)$ on chain complexes. Then $$\ker c_\# \hookrightarrow C_0(X) \stackrel{c\#}{\to} C_0(*)$$ is trivially exact. But by the definitions of $c_\#$ and $\epsilon$ we have $\ker c_\# = \ker \epsilon$, thus $$\ker \epsilon \stackrel{\mu}{\to} C_0(X) \stackrel{c\#}{\to} C_0(*)$$ is exact. The induced $c_* : H_0(X) \to H_0(*)$ is given by $$c_*([\xi]) = [c_\#(\xi)] . $$ Thus $[\xi] \in \ker c_*$ means $[c_\#(\xi)] = 0$. Since the quotient map $C_0(*) \to H_0(*)$ is an isomorphism, the latter is equivalent to $c_\#(\xi) = 0$ and therefore equivalent to $\xi \in \text{im} \mu$ which is the same as $[\xi] = [\mu(\eta)] = \iota([\eta]) \in \text{im} \iota$.

Therefore $\ker c_* = \text{im} \iota$ as desired.

Paul Frost
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