Lemma 2.3: Let $M$ be a an $n-$manifold and $G$ be an abelian group.
(a) For any compact $K \subset M$ and $i > n$ $$H_i(M,M-K;G) = 0$$
(b) If $u \in H_n(M,M-K;G)$ and $\rho_x(u) = 0$ for all $x\in K$, then
$u=0$.
I'd like to understand the case $4$ of the Lemma, i.e $M = \mathbb{R}^n$ and $K$ an arbitrary compact.
$\bullet$ We use the following fact which I'm unable to prove: "We assert that for any $u \in H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus K)$ exists an open set $N$ containing $K$ and elements $u'\in H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus N)$ such that $k_*(u') = u$ where $k$ is the inclusion map.
I do understand the "recall" part of the sentence, but I don't understand how to explicitly write the details.
Let $(X,A) \subseteq (\mathbb{R}^n,\mathbb{R}^n \setminus K)$ with $v \in H_i(X,A)\subseteq H_i(\mathbb{R}^n,A)\longmapsto u \in H_i(\mathbb{R}^n,\mathbb{R}^n \setminus K)$ under inclusion be given.
What I thought : Since $A \subseteq \mathbb{R}^n \setminus K$ and both $A,K$ are compact, we have "positive distance". Being $K$ compact, we can choose $N$ as finite union of open balls $B_j$ that cover $K\subseteq \bigcup\limits_{j=1}^n B_j$ such that $B_j \cap A = \varnothing$ ($N \cap A = \varnothing$).
So in reality $v \in H_i(\mathbb{R}^n,\mathbb{R}^n \setminus N) \longmapsto u \in H_i(\mathbb{R}^n,\mathbb{R}^n \setminus K)$.
Is this reasoning correct? I'm not sure about last sentence since I think we could have that $v$ has image in $\text{Imm} \mathbb{R}^n \setminus N$.
Best Answer
It is known that singular homology has compact carriers which means that for each space $X$ and each $x \in H_i(X)$ there exists a compact $C \subset X$ such that $x$ is in the image of the inclusion induced $j_* : H_i(C) \to H_i(X)$ (let me know if you want a proof). The same is of course true for reduced homology groups (which only differ in dimension $0$ from the unreduced groups).
Le $N$ be any set such that $K \subset N \subset \mathbb R^n$. Now consider the long exact sequences of reduced homology groups of the pairs $(\mathbb{R}^{n},\mathbb{R}^{n}\setminus K)$ and $(\mathbb{R}^{n},\mathbb{R}^{n}\setminus N)$. Since all reduced homology groups of $\mathbb R^n$ are $0$, we get for $i > 0$ commutative squares
$\require{AMScd}$
$$\begin{CD} H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus N) @>\partial>>\tilde H_{i-1}(\mathbb{R}^{n}\setminus N)\\ @Vk_*VV@Vj_*VV@.@.\\ H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus K) @>\partial>>\tilde H_{i-1}(\mathbb{R}^{n}\setminus K) \end{CD}$$ where the horizontal arrows are isomorphisms and the vertical arrows are inclusion induced. Note that the long exact sequences show that $H_0(\mathbb{R}^{n},\mathbb{R}^{n}\setminus N) = H_0(\mathbb{R}^{n},\mathbb{R}^{n}\setminus K) = 0$, thus your question concerning the existence of $N$ and $u'$ is trivial for $i = 0$ (you may take $N = \mathbb{R}^{n}$).
Thus, given the element $u$, consider $x = \partial(u)$ and find compact $C \subset \mathbb{R}^{n}\setminus K$ and $x' \in H_{i-1}(C)$ such that $j_*(x') = x$. Clearly $N = \mathbb R^n \setminus C$ is an open set containing $K$. By definition $C= \mathbb{R}^{n}\setminus N$. Now let $u' = \partial^{-1}(x')$.