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Since the 0 homology of a line with two vertices is $\mathbb{Z}$, may I please have an example of two vertices not filled in by a line ? I just can not figure it out - thank you!!!!

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Your question does not seem to be well posed, but I try my best to answer it. Please correct me if I got you wrong.

For the homology of a line(-segment), i.e. the unit interval $[0,1]$, note that $[0,1]$ is contractible, i.e. homotopy equivalent to a point. By homotopy invariance of singular homology we thus have $H_\bullet([0,1]) \cong H_\bullet(*)$ and in particular $H_0([0,1]) \cong H_0(*) = \mathbb Z$.

For the homology of two separate points, i.e. the topological space given by the disjoint union $* \sqcup *$ we may use the fact that singular homology satisfies what is called the additivity axiom, which gives us $H_\bullet(* \sqcup *) \cong H_\bullet(*) \oplus H_\bullet(*)$ and in particular $H_0(* \sqcup *) \cong H_0(*) \oplus H_0(*)\cong \mathbb Z \oplus \mathbb Z$.

Addendum Let us try to make the calculations align with the „counting holes“ slogan.

Recall that for a positive integer $n \in \mathbb N$ one has $H_n(\mathbb S^n) = \mathbb Z$ and $H_0(\mathbb S^n) = \mathbb Z$. The $n$th homology thus kind of tells us that we have one $n-1$-dimensional hole (the interior of the sphere). You may have heard that the $0$-th homology of a space counts its path-components. So as $\mathbb S^n$ is connected this kind of explains, why we have $\mathbb Z$ there aswell.

It is important to note that the line segment $[0,1]$ is not the $0$-dimensional sphere, but the $1$-dimensional disk. Hence we expect $H_1([0,1]) = 0$ because there are no $0$-dimensional holes and $H_0([0,1]) = \mathbb Z$ since $[0,1]$ has precisely one path connected component.

Finally for the real $0$-dimensional sphere, which is the boundary of the $1$-dimensional disk, i.e. consists of two disjoint points the intuition of counting wholes kind of breaks down. I mean, what is a $-1$-dimensional hole supposed to be? The counting connected components thing still works though, which might explain why $H_0(\mathbb S^0) = \mathbb Z^2$.

Jonas Linssen
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  • Here is what I mean: a circle has $\mathsf{H}{1} = \mathbb{Z}$ because it encircles a hole - there is a gap between the lines. $\mathsf{H}{0}$ of a line also equals $\mathbb{Z}$. There isn't a visible gap between two points to me. What should I do? –  Jan 20 '20 at 02:22
  • Do I count one vertex twice and see if there is a line between it? A circle seems like that to me. –  Jan 20 '20 at 02:48
  • I edited my answer. Please tell me if it helped. – Jonas Linssen Jan 20 '20 at 08:25
  • What do you mean with the disk is out? – Jonas Linssen Jan 20 '20 at 11:14
  • In the example $\mathbb{S}^{; 1}$, $\mathsf{H}{;1}$ tracks a missing $D^{; 2}$. In the example $D^{; 1}$ $\mathsf{H}{;1}$ tracks a missing $D^{; 0}$ instead? –  Jan 20 '20 at 11:39
  • Maybe it helps to think of $\mathbb S^1$ as $\mathbb R^2\setminus {0}$ ie. as the punctured plane, which we can do by homotopy equivalence. In this low dimensional case a hole is therefore not about missing some connecting lines between antipodal points as you put it, but rather having a missing point in the plane. This is what I meant when saying the hole is 0-dimensional. Hence when talking about $\mathbb S^0$ you can think of the hole being -1-dimensional, which is nonsense to me. – Jonas Linssen Jan 20 '20 at 11:50
  • Thank you ! -
    • –  Jan 20 '20 at 11:59
  • please wait- if $\mathsf{H}{; 1} ; ( ; \mathbb{S}^{; 1} ; )$ is 0 dimensional by that homotopy equivalence, I don't see how $\mathsf{H}{; 2} ; ( ; \mathbb{S}^{; 2} ; )$ is 1 dimensional- can you please tell me ^_^.......... –  Jan 20 '20 at 12:24
  • when I make $\mathbb{S}^{; 2} ; \simeq ; \mathsf{R}^{; 3} ;$ \ $; { ; 0 ; }$ then I get the same thing as you put for $\mathbb{S}^{; 1}$, a 0 dimension hole. –  Jan 20 '20 at 12:38
  • Well I just noticed there is a flaw somewhere in my explanation. Let me think about it... – Jonas Linssen Jan 20 '20 at 12:55
  • I think you got me there, the hint with just considering the pointed space was misleading at best. I am afraid, my intuition is wrong and my answer does not make sense at all, as it did rely on the dimension of a hole being less than the dimension of the sphere, which now appears to be rubbish to me. I apologize. Please feel free to uncheck my answer. – Jonas Linssen Jan 20 '20 at 16:57
  • The line - disc D^1 is the only 1 cell, therefore if it were labelled as $a$, $a ; - ; a$ would be the only cycle, so therefore $\mathsf{H} ; ( ; 1; )$ is 0. Does that sound better? –  Jan 21 '20 at 09:23