Lemma 6.3 in Vick Can anyone help me understand the proof of lemma 6.3 in the book by Vick on Homology Theory: an Introduction to Algebraic Topology.


I have no idea what Vick tries to do here. I follow that showing $i_*(b')=0$. But then he creates this weird construction with the cubes.
The first thing I do not understand is why the image of $b'$ vanishes under the morphism $$H_{n-1}(V)\to H_{n-1}(\mathbb{R}^n\setminus\{p\})$$ I see why this holds for $p\in \mathbb{R}^n\setminus U$, but Vick takes $p\in \overline{K}$, which is not disjoint from $U$.
The second thing I do not understand is why the induced morphisms from the inclusions $$H_{n-1}(Q_{k+1})\to H_{n-1}(Q_{k}),\quad H_{n-1}(Q_{k+1})\to H_{n-1}(\mathbb{R}^n\setminus P_{k+1})$$ send $b'$ to zero. I suppose this is why Vick chose points in $\overline{K}$ and cubes that cover it. But the intuition as well as the technical details are completely lost on me.
 A: Note that $Q\cap U$ is an open subset of $Q$, so $K=Q\backslash (Q\cap U)$ is a closed subset of $Q$. Hence, $\overline K=K$, here closure is taken w.r.t $Q$.
Choose a point $p\in  \overline K$ and let us rename the different maps in the commutative diagram:
$(1)$ The inclusion induced map $i_*\colon H_{n-1}(V)\to H_{n-1}(U)$ which sends $b'$ to $b$.
$(2)$ The isomorphism $\Delta_1\colon H_{n}(\Bbb R^n, U)\to H_{n-1}(U)$ with $\Delta_1^{-1}(b)=a$.
$(3)$ The map $j_{p*}\colon H_n(\Bbb R^n,U)\to H_n(\Bbb R^n,\Bbb R^n\backslash p)$ sends $a$ to $0$ from the hypothesis of the lemma. Note that the hypothesis says for all $p\in \Bbb R^n\backslash U$.
$(4)$ The isomorphism $\Delta_2\colon H_n(\Bbb R^n,\Bbb R^n\backslash p)\to H_{n-1}(\Bbb R^n\backslash p)$ sends $0$ to $0$.
$(5)$ So, the inclusion induced map $\ell_*\colon H_{n-1}(U)\to H_{n-1}(\Bbb R^n\backslash p)$ sends $b$ to $0$ as the right square is commutative. That is $\ell_*(b)=\Delta_2\circ j_{p*}\circ \Delta_1^{-1}(b)=0$.
$(6)$ Hence, the inclusion induced map $k_*\colon H_{n-1}(V)\to H_n(\Bbb R^n\backslash p) $ sends $b'$ to zero as the left square is commutative. That is $k_*(b')=\text{Id}\circ k_*(b')=\ell_*\circ i_*(b')=\ell'(b)=0$. Here, $\text{Id}\colon H_n(\Bbb R^n\backslash p)\to H_n(\Bbb R^n\backslash p)$ is induced from identity map of $\Bbb R^n\backslash p$.

$(7)$ Therefore, the inclusion induced homomorphism $H_{n-1}(V)\to H_n(\Bbb
 R^n\backslash p)$ sends $b'$ to $0$ for all $p\in  \overline K$.


The Induction Hypothesis is: the image of $b'$ under the inclusion induced map $H_{n-1}(V)\to H_{n-1}\big(Q\backslash (P_1\cup\cdots \cup P_k)\big)$ is zero. Note that $Q_k=Q\backslash (P_1\cup\cdots \cup P_k)$.

Let  $p\in \text{int}(P_{k+1})\cap \overline K$. Let $B(p,r):=\{x\in \Bbb R^n:||x-p||<r\}\supseteq P_{k+1}$ for some $r>0$. Then, there are strong deformation retracts of $\Bbb R^n\backslash P_{k+1}$ and $\Bbb R^n\backslash p$ onto $\partial B(p,r)$ via radial projection w.r.t. $p$.
Hence, the inclusion induced maps $H_{n-1}\big(\partial B(p,r)\big)\xrightarrow{\alpha_*} H_{n-1}(\Bbb R^n\backslash P_{k+1})$ and $H_{n-1}\big(\partial B(p,r)\big)\xrightarrow{\beta_*} H_{n-1}(\Bbb R^n\backslash p) $ are isomorphisms.
But, $\beta_*=H_{n-1}\big(\partial B(p,r)\big)\xrightarrow{\alpha_*} H_{n-1}(\Bbb R^n\backslash P_{k+1})\xrightarrow{\text{inclusion}_*}H_{n-1}(\Bbb R^n\backslash p)$.  In other words, the inclusion induced map $H_{n-1}(\Bbb R^n\backslash P_{k+1})\to H_{n-1}(\Bbb R^n\backslash p)$ is an isomorphism.
Now, the composition of inclusion induced maps $H_{n-1}(V)\to H_{n-1}(\Bbb R^n\backslash P_{k+1})\to H_{n-1}(\Bbb R^n\backslash p)$ is the inclusion induced map $H_{n-1}(V)\to H_{n-1}(\Bbb R^n\backslash p)$, which sends $b'$ to $0$ by $(7)$ as $p\in \text{int}(P_{k+1})\cap \overline K\subseteq  \overline K$.
Therefore, the inclusion induced map $H_{n-1}(V)\to H_{n-1}(\Bbb R^n\backslash P_{k+1})$ sends $b'$ to $0$ as the inclusion induced map $H_{n-1}(\Bbb R^n\backslash P_{k+1})\to H_{n-1}(\Bbb R^n\backslash p)$ is an isomorphism.
