Understanding a part of the proof of the Invariance of domain from Hatcher

Question

On page number 126 of Allen Hatcher's Algebraic Topology, he has proved invariance of domain.

In the second line of his proof he writes,

From the long exact sequence for the pair $(\Bbb R^m, \Bbb R^m - \{x\})$ we get $H_k(\Bbb R^m,\Bbb R^m - \{x\}) \approx \tilde{H}_{k-1}(\Bbb R^m - \{x\}).$

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But from long exact sequence we have $H_k(\Bbb R^m,\Bbb R^m - \{x\}) \approx H_{k-1}(\Bbb R^m - \{x\})$ for all $k \gt 0$ and we know that $\tilde{H}_{k-1}(\Bbb R^m - \{x\}) \approx H_{k-1}(\Bbb R^m - \{x\})$ for all $k \gt 1$ with $H_0(\Bbb R^m - \{x\}) \approx \tilde{H}_0(\Bbb R^m - \{x\}) \oplus \Bbb Z$.

So my question is that how does it particularly follow that $H_1(\Bbb R^m,\Bbb R^m - \{x\}) \approx \tilde{H}_{0}(\Bbb R^m - \{x\})$ ? Am I missing something?

Answer 1

The long exact sequence for the pair $(A,B)$ with reduced homology is $$\cdots\to \widetilde{H}_{k+1}(A) \to H_{k+1}(A,B) \to \widetilde{H}_{k}(B) \to \widetilde{H}_{k}(A)\to \cdots$$ which is mentioned on page 118.

Since $A=\mathbb{R}^m$ is contractible, $\widetilde{H}_{k}(A)=0$ for all $k$, hence there are isomorphisms $H_{k+1}(\mathbb{R}^m,\mathbb{R}^m-\{x\})\to \widetilde{H}_{k}(\mathbb{R}^m-\{x\})$ for all $k$.

A way to think about what's going on with respect to unreduced homology is that at the end of the sequence $$\cdots\to H_{1}(A,B) \to H_0(B) \to H_0(A) \to H_0(A,B)\to 0$$ to recall that $H_0$ is measuring the number of connected components. We have $H_0(A)=\mathbb{Z}$ since $A$ has a single connected component. In the case $m>1$, then $B$ has a single connected component as well, and $H_0(B)\to H_0(A)$ is an isomorphism, and one can deduce that $H_1(A,B)=0$. In the case $m=1$, then $B$ has two connected components, and the map $H_0(B) \to H_0(A)$ is $\mathbb{Z}\oplus\mathbb{Z}\to \mathbb{Z}$, defined by $(a,b)\mapsto a+b$. The kernel of this is generated by $(1,-1)$, which is a copy of $\mathbb{Z}$, and so $H_1(A,B)\approx \mathbb{Z}$ in this case.

Answer 2

Well, you can use the reduced exact sequence for a pair.

let $A=\mathbb R^m \setminus \{x\}$

$\tilde H_1(\mathbb R^m) \to H_1(\mathbb R^m,A) \to \tilde{H_0}(A) \to \tilde H_0(\mathbb R^m) \to H_0(\mathbb R^m,A) \to 0$

so in particular, we have

$0 \to H_1(\mathbb R^m,A) \to \tilde H_0(A) \to 0 $

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Regarding your error, you can apply the usual sequence to get that

$$0 \to H_1(\mathbb R^m,A) \to H_0(A) \to \mathbf Z \to 0.$$

There are some technical remarks needed to still get the result, but if you assume some form of splitting, then you get $H_0(A)=H_1(\mathbb R^m,A) \oplus \mathbf Z$,