# Homology relative to a point is reduced homology

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!

• We have $H_1 (*) = 0$. I think you mean $H_0 (*) = \mathbb{Z}$. Mar 27, 2020 at 1:15
• In time, we have that $\widetilde{H}_n (*) = 0$, $\forall n$, and, in general, $H_0(X) = \widetilde{H}_0 (X) \oplus \mathbb{Z}$. Mar 27, 2020 at 1:20
• @Gustavo Yes, thank you, I will correct this typo now. Mar 27, 2020 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

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

• 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? Mar 27, 2020 at 1:46
• @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 $x$. Mar 27, 2020 at 14:17
• 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 $1$). Mar 27, 2020 at 14:18