Proof of $\tilde{H}_*(X) \cong H_*(X, x_0) $.

Question

Let $X$ be a topological space and $x_0 \in X$. I want to show that the relative homology group, $H_*(X, x_0)$, is isomorphic to the reduced homology group, $\tilde{H}_*(X)$. By considering the long exact sequence of $(X, x_0)$ we get $$H_k(x_0) = 0 \to H_k(X) \to H_k(X, x_0) \to H_{k-1}(x_0) \to \ldots$$ For $k \ge 2$, $H_{k-1}(x_0) = 0$ so that $H_k(X, x_0) \cong H_k(X) = \tilde{H}_k(X).$ Now my problem is for $k = 1, 0$.

For $k = 1$, the above sequence becomes $$ 0 \to H_1(X) \xrightarrow{j} H_1(X, x_0) \xrightarrow{\delta} H_{0}(x_0) \cong \mathbb Z,$$ so that $H_1(X) \cong \text{im}(j) = \text{ker}(\delta)$, how can I see that $\delta$ is trivial ?

For $k = 0$, I get $$H_{0}(x_0) \cong \mathbb Z \xrightarrow{i} H_0(X) \xrightarrow{j} H_0(X, x_0).$$ By using that $\delta$ is trivial, we have $\text{ker}(i) = 0$ and $H_0(X) \cong \text{im}(j)$. I do not see how I should get the result..

Any help would be greatly appreciated.

Answer 1

Recall that $H_0(X)$ is the free abelian group generated by the path-connected components of $X$. That is, if $\{X_i\}_{i\in I}$ are the path-connected components of $X$, then $H_0(X) \cong \bigoplus_{i\in I}\mathbb{Z}$. The map $\{x_0\} \to X$ induces the map $H_0(\{x_0\}) \to H_0(X)$ which sends the generator of $H_0(\{x_0\}) \cong \mathbb{Z}$ to the element of $\bigoplus_{i\in I}\mathbb{Z}$ which has a $1$ in the index corresponding the path-connected component of $X$ containing $x_0$, and $0$ in every other index. This is the map $i$ and it is injective. In general, if $A \subset X$ and every path-connected component of $A$ is contained in a path-connected component of $X$, then the induced map $H_0(A) \to H_0(X)$ is injective. The only way the map can fail to be injective is if two path-connected components of $A$ are contained in the same path-connected component of $X$.

Answer 2

You can make explicit the connecting morhpism $\delta$ following the steps in the proof of the long exact sequence lemma. Take an element that is a border $[\alpha]\in \frac{C_1(X)}{C_1(x_0)}$, this comes from $\alpha\in C_1(X)$ whose is such that $\partial(\alpha)\in C_0(x_0)$. Then $\partial(\alpha)=x_0-x_0=0\in C_0(X)$. Because the map $C_0(x_0)\to C_0(X)$ is injective this comes from $0$ in $C_0(x_0)$. This implies $\delta$ is the zero map.

Note that in $\partial(\alpha)=x_0-x_0=0\in C_0(X)$ I treated $\alpha$ like a generator and not a linear combination, but in any case $\partial(\alpha)=0$ holds.

Answer 3

You do not tell us which definition of reduced homology groups you want to use. There are two two approaches:

1. Define them as the homology groups of the augmented singular chain complex.

2. Define them as $\tilde H_k(X) =\ker(c_* : H_k(X) \to H_k(*))$, where $c : X \to *$ is the unique map onto a one-point space $*$.

Both definitions are equivalent, but this requires a proof.

Anyway, the inclusion map $i : \{x_0\} \hookrightarrow X$ has the unique map $c : X \to \{x_0\}$ as a left inverse (i.e. $c i = id$), thus we get $c_*i_* = id$ which shows that $i_* : H_0(x_0) \to H_0(X)$ is injective. That is, we have $\ker i_* = 0$ . The long exact sequence shows then that $\partial : H_1(X, x_0)\to H_{0}(x_0)$ has image $0$. Hence $j : H_1(X) \to H_1(X,x_0)$ is an isomorphism.

Next we get a short exact sequence

$$0 \to H_0(x_0) \stackrel{i_*}{\to} H_0(X) \stackrel{j}{\to} H_0(X,x_0) \to 0$$ This sequence splits since $i_*$ has $c_*$ as a left inverse. Thus $$\phi : H_0(X) \to H_0(x_0) \oplus H_0(X,x_0), \phi(g ) = (c_*(g),j(g))$$
is an isomorphism. Letting $p_1 : H_0(x_0) \oplus H_0(X,x_0) \to H_0(x_0)$ denote the projection onto the first summand, we see that $c_* = p_1 \phi$. Thus $\ker c_* \cong \ker p_1 = 0 \oplus H_0(X,x_0) \cong H_0(X,x_0)$. This gives a proof for the second definition.

If you want to use definition 1, you need some additional ingredients.