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

  • $\begingroup$ 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. $\endgroup$
    – Paul Frost
    Jul 29, 2021 at 9:38
  • $\begingroup$ @PaulFrost I don't understand how to create $N$ properly and verify that satisfies the requests $\endgroup$ Jul 29, 2021 at 9:42
  • $\begingroup$ 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? $\endgroup$
    – Paul Frost
    Jul 29, 2021 at 9:47
  • $\begingroup$ @PaulFrost Am I allowed to cite the words? I think i'm not suppose to scan or add a picture $\endgroup$ Jul 29, 2021 at 9:48
  • $\begingroup$ 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 $\endgroup$
    – Paul Frost
    Jul 29, 2021 at 9:51

1 Answer 1


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


$$\begin{CD} H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus N) @>\partial>>\tilde H_{i-1}(\mathbb{R}^{n}\setminus N)\\ @Vk_*VV@Vj_*VV@.@.\\ H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus K) @>\partial>>\tilde H_{i-1}(\mathbb{R}^{n}\setminus K) \end{CD}$$ 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 $u' = \partial^{-1}(x')$.

  • $\begingroup$ 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. $\endgroup$ Sep 12, 2021 at 15:39
  • $\begingroup$ @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. $\endgroup$
    – Paul Frost
    Sep 13, 2021 at 7:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .