# Proof of Lemma 2.3 p.354 Massey

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 $$\text{Imm} \mathbb{R}^n \setminus N$$.

• I do not understand what you ask. You have an inclusion $j : (X,A) \hookrightarrow (\mathbb{R}^n,\mathbb{R}^n \setminus K)$ and it induces $j_*$ on homology groups. Jul 29, 2021 at 9:38
• @PaulFrost I don't understand how to create $N$ properly and verify that satisfies the requests Jul 29, 2021 at 9:42
• You should completely state Massey's Lemma. Not everybody has access to his book. Does he introduce $N$ and what does he claim about it? Jul 29, 2021 at 9:47
• @PaulFrost Am I allowed to cite the words? I think i'm not suppose to scan or add a picture Jul 29, 2021 at 9:48
• In my opinion you are allowed to quote. But perhaps you should ask a question on Meta concerning the rules about quoting resp. embedding scanned parts of books (copyright!). math.meta.stackexchange.com Jul 29, 2021 at 9:51

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 $$u' = \partial^{-1}(x')$$.
• Are you able to show that singular homology has compact carriers? In particular the "b part" of the theorem, i.e that if a homology class is zero under some inclusion of some compact in the space then exist a compact in between the previous two such that the class already was $0$ when included in new compact? Any other answer to my unanswered question is appreciated. Sep 12, 2021 at 15:39
• @jacopoburelli Concerning compact carriers: The image of a singular simplex is a compact subspace of $X$. Thus a singular chain factors through the union of the images of its singular simplices, which is a compact subspace of $X$. For part b one has to know the definition of $\rho_x$. I do not have access to Massey's book. Sep 13, 2021 at 7:57