I am trying to prove that $H_n(X, *) \cong \widetilde{H_n}(X)$ for all $n$. Hatcher uses the long exact sequence of reduced homology groups to prove this for all $n$ (example 2.18), but my professor has not mentioned this sequence. I am looking for guidance on the case $n=0$, as I am quite stuck. I would also appreciate feedback on my attempt for $n \geq 1$:

Let $n \geq 1$. Consider the long exact sequence of relative homology groups. Since $H_n(*) \cong 0$ and $\widetilde{H_n}(X) \cong H_n(X)$, we rewrite the sequence as $$ \cdots \xrightarrow{j_*} H_{n+1}(X, *) \xrightarrow{\partial_*} 0 \xrightarrow{i_*} \widetilde{H_n}(X) \xrightarrow{j_*} H_n(X, *) \xrightarrow{\partial_*} 0 \xrightarrow{i_*}\cdots .$$ By exactness, $\text{Ker } j_* = \text{Im } i_* = 0$ and $\text{Im }j_* = \text{Ker }\partial_* = H_n(X, *)$. Thus $j_* : \widetilde{H_n}(X) \to H_n(X, *)$ is an isomorphism.

This seems almost correct, but I think there is an issue in the case $n = 1$, since $H_0(*) = \mathbb{Z} \neq 0$, and thus $\text{Ker }\partial_* \neq H_1(X, *)$. How I can I fix this?

Thank you!

  • 1
    $\begingroup$ We have $H_1 (*) = 0$. I think you mean $H_0 (*) = \mathbb{Z}$. $\endgroup$
    – Gustavo
    Mar 27 '20 at 1:15
  • $\begingroup$ In time, we have that $\widetilde{H}_n (*) = 0$, $\forall n$, and, in general, $H_0(X) = \widetilde{H}_0 (X) \oplus \mathbb{Z}$. $\endgroup$
    – Gustavo
    Mar 27 '20 at 1:20
  • $\begingroup$ @Gustavo Yes, thank you, I will correct this typo now. $\endgroup$
    – Victor
    Mar 27 '20 at 1:38

The definition of the reduced homology group is as follows. Any space $X$ admits a unique map $X\rightarrow\ast$ to the one-point space and we set

$$\widetilde H_n(X)=\ker\left(H_n(X)\rightarrow H_n(\ast)\right).$$

Assuming that $X$ is nonempty, any choice of point $x\in X$ defines a map $x:\ast\rightarrow X$ which splits the surjection $X\rightarrow\ast$. Then by the functorality of homology, the induced map $x_*:H_n(\ast)\rightarrow H_n(X)$ is injective.

Thus when we consider the long exact sequence

$$\dots\rightarrow H_n(\ast)\xrightarrow{x_*} H_n(X)\rightarrow H_n(X,\ast)\rightarrow H_{n-1}(\ast)$$

we see by exactness that it splits in each degree to give

$$H_n(X)\cong H_n(X,\ast)\oplus H_n(\ast).$$

But under this isomorphism the map $H_n(X)\rightarrow H_n(\ast)$ becomes the projection onto the second factor. Hence

$$\ker\left(H_n(X)\rightarrow H_n(\ast)\right)\cong \ker\left(H_n(X,\ast)\oplus H_n(\ast)\xrightarrow{pr_2} H_n(\ast)\right)$$

and thus

$$\widetilde H_n(X)\cong H_n(X,\ast).$$

  • $\begingroup$ I'm afraid this is a bit over my head. I am using the definition of reduced homology groups found in Hatcher, obtained by augmenting the chain complex with a map $\epsilon : C_0(X) \to \mathbb{Z}$. It is not apparent to me why your definition is equivalent. Furthermore, I am not familiar with the term "splitting." Is there a more elementary (but perhaps longer) way to approach this problem? $\endgroup$
    – Victor
    Mar 27 '20 at 1:46
  • $\begingroup$ @Victor, I think you'll find that this is the most elementary approach (barring a few words here and there). I only use functorality and exactness (making no use of chains, so in particular it applies to any homology theory, not just singular (see Hatcher $\S$ 2.3)). Firstly, by 'splitting', I mean a right inverse: the one-point space retracts off of any nonempty space $X$ (i.e. the composite $\ast\xrightarrow{x}X\rightarrow\ast$ is equal to the identity. Here $x\in X$ is a point, and $x:\ast\rightarrow X$ is the map which sends the unique point to $x$. $\endgroup$
    – Tyrone
    Mar 27 '20 at 14:17
  • $\begingroup$ The definition of reduced homology that I am using is that given by Hatcher in $\S$ 2.3. The augmentation $\epsilon:C_0(X)\rightarrow \mathbb{Z}$ is the zeroth component of the chain map induced by $X\rightarrow \ast$, after identifying $C_0(X)\cong\mathbb{Z}$ (there is a unique map $\Delta^0\rightarrow \ast$, so $C_0(X)$ is free abelian of rank $1$). $\endgroup$
    – Tyrone
    Mar 27 '20 at 14:18

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