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We have a long exact sequence

$...\to H_0(A)\to H_0(X)\to H_0(X,A)\to 0$

If $H_0(X,A)=0$, then by exactness at $H_0(X)$: Image $(H_0(A)\to H_0(X))=H_0(X)=$ ker $(H_0(X)\to H_0(X,A))$ so the map $i_*:H_0(A)\to H_0(X)$ is surjective.

Let $X_a$ be a path component of $X$. Suppose on the contrary that $A\cap X_a=\emptyset$. I don't understand how to get contradiction from here and how to prove the converse. Please help. Thanks.

The linked post does not answer my question. The linked post does not clarify why A intersects every path component of $X$. The answer here Path-components and relative homology does not explain the steps so I don't understand it.

I don't understand the answer here either, it seems too complicated $H_0(X,A) = 0 \iff A$ meets each path-component of $X$. and same for this $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$.

None of the linked posts explains how to use surjectivity of $i_*$ to show that each path component of X intersects $A$.

Koro
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  • Let $x_a\in X_a$. Then it generates a cyclic subgroup of $H_0(X)$. It is the image of no element of $H_0(A)$, as in order to be so, there must be a path from $x_a$ to some element of $A$. – Ted Shifrin Mar 26 '23 at 03:36
  • See for example https://math.stackexchange.com/q/2963093 and the links in this post. – Paul Frost Mar 26 '23 at 08:09
  • @PaulFrost: no, it does not answer my question. The linked post does not talk about pathcomponents of X-neither the post nor the posted answer there. – Koro Mar 26 '23 at 10:08
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    @Koro That's not quite true. The linked post mentions the question you have and further links to other duplicates, as Paul said. See here and here – FShrike Mar 26 '23 at 10:25
  • @FShrike: But my question is not answered by any of the posts linked. I don't understand how they prove A meets all path connected components of X. – Koro Mar 26 '23 at 13:34
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    I don't know myself. But both ask that exact question, show some working, and have the working completed by the answerer. It seems to me that these are indeed answering the question. To get this reopened as a non duplicated you could maybe take some extracts from those links and say why you don't understand them – FShrike Mar 26 '23 at 15:59
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    Provide more detail about, in particular, how the answer at https://math.stackexchange.com/a/61971/42781 is not sufficient. – John Palmieri Mar 26 '23 at 16:53
  • @JohnPalmieri: I don't understand how they write the direct sum of $H_0(X, A) $. – Koro Mar 27 '23 at 03:20
  • Are you using Hatcher's book? It's Proposition 2.6 but for relative homology. (The same proof works for relative homology as for absolute homology.) – John Palmieri Mar 27 '23 at 03:49
  • @JohnPalmieri: Yes, but only some sections. I have already explained my issue with the answer: I don't understand how they use surjectivity of $H_0(A)\to H_0(X)$ (I get this surjectivity using long exact sequence) to conclude path components of X intersect A. – Koro Mar 27 '23 at 05:57
  • The sentence "If $A$ is empty, then exactness immediately tells you that $H_0(X,A)$ is non-zero" is the contrapositive of the argument you're asking about. ("$A$ empty implies $H_0(X,A)\neq 0$" is equivalent to "$H_0(X,A) = 0$ implies $A$ non-empty.") What part of that do you not understand? – John Palmieri Mar 27 '23 at 15:49
  • @JohnPalmieri: I understand this but I don't understand the last para in the answer at all. It seems to be a series of claims with no explanation at all. It's certainly not written for a beginner like me. – Koro Mar 28 '23 at 03:43
  • Through this post, I was expecting some beginner friendly answer that would make sense to me but the question has been closed. There are other solutions available online (outside this site) but they make no sense to me. :( – Koro Mar 28 '23 at 03:49

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