First, this is the link to the book, for convenience: https://pi.math.cornell.edu/~hatcher/AT/AT.pdf#page=246
Lemma 3.27. Let $M$ be an $n$-manifold and let $A \subset M$ be a compact subset. Then:
(a) If $x \mapsto \alpha_x$ is a section of the covering space $M_R \to M$, then there is a unique class $\alpha_A \in H_n(M,M-A;R)$ whose image in $H_n(M,M-x;R)$ is $\alpha_x$ for all $x \in A$. (The construction of $M_R$ is described in above the lemma.)
(b) $H_i(M,M-A;R)=0$ for $i>n$.
The proof consists of four steps, and I got stuck at the last sentence of the second step, and the third step.
- The last sentence in the second step: This sentence is essentially equivalent to the following:
Suppose the lemma holds when $M= \Bbb R^n$. Then the lemma holds for arbitrary $M$ and $A$, if $A \subseteq \Bbb R^n \subseteq M$(i.e., $A$ is contained in an open neighborhood in $M$ which is homeomorphic to $\Bbb R^n$, and which we just identify with $\Bbb R^n$), by excision.
I know that by excision we have $H_i(M,M-A;R) \cong H_i(\Bbb R^n, \Bbb R^n-A;R)$, so (b) obviously follows from the case $M=\Bbb R^n$. But I cannot see how (a) follows from the case $M=\Bbb R^n$.
- The last sentence in the third step: This sentence just says that the lemma holds if $M=\Bbb R^n$ and $A$ is a compact convex subset, because for $x \in A$, both $\Bbb R^n-A$ and $\Bbb R^n-x$ deformation retract onto a sphere centered at $x$.
I see that the spaces $\Bbb R^n-A$ and $\Bbb R^n-x$ indeed deformation retracts onto a sphere, so (b) clearly follows. But again I can't see how does this imply (a).
Thus, in summary, I can't see how (a) holds, in steps (2) and (3) in the proof. How do I have to show (a)?
Thanks in advance.
Best Answer
I think you might find my answer useful. You understood how in (2) he reduced to the case that $A$ is a subset of a neighborhood in $M$ homeomorphic to $\mathbb{R}^n$. What is important to remember that excision isomorphism is induced by inclusion.
For each $x\in A$ we can take the representative of $\alpha_x$ to be an element of $H_n(\mathbb{R}^n,\mathbb{R}^n-\{x\};R)$. By assumption there is a unique class $\alpha_A$ in $H_n(\mathbb{R}^n,\mathbb{R}^n-A;R)$, such that $i_*(\alpha_A)=\alpha_x$ for all $x \in A$. Take $\alpha_A$ to be your class in $H_n((M,M-A);R)$.
In (3) choose $x \in A$. As hatcher explains will have an isomorphism $f$ between the relevant homology groups so we can take $\alpha_A$ to be $f(\alpha_x)$. Note that $\alpha_x =r \in R$ for all $x \in A$ else the image of $\alpha$ would be disconnected a contradiction to the fact that $A$ is convex hence connected. Thus $\alpha_A$ is the desired element.