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$H_{1}$ of a line segment is $\mathsf{Z}^{e - v + 1}$, so $\mathsf{Z}^{1 - 2 + 1}$ or $\mathsf{Z}^{0}$ \footnote{nj wildberger, homology video}... That equals 0 right? I have only taken calulus III, not linear algebra yet. With two disjoint line segments I get $\mathsf{Z}^{2-4+1 \; = \; - 1}$.... How am I actually supposed to extend that definition to the two disjoint line segments - because, I can't compute what $\mathsf{Z}^{-1}$ is.

gt6989b
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    If it can be continuously deformed to a point (just “press together” the extremities of the line), the $H_1$ is zero. If your space is the disjoint reunion of two subspaces (so that there isn’t a path from one to the other), the total $H_1$ is the sum of the $H_1$ of the subspaces. – Aphelli Jan 19 '20 at 16:14

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Yes, a line segment has trivial homology. TWO line segments will have nontrivial 0 homology, but trivial homology in every other dimension.

A quick way to see this is by "contracting" the lines down to points, which if you have not seen yet, you will probably see soon. Loosely, homology measures how many "holes" your space has, where a 0 dimensional hole is two points not filled in by a line, a 1 dimensional hole is a circle that isn't filled in by a disk, a 2 dimensional hole is a sphere not filled in by a ball, and so on.

In the one line case, you have every point connected to every other point, and so we get trivial homology (with homology is $\mathbb{Z}^1$ in dimension 0, because there is 1 connected component.

In the two line case, we have TWO connected components, so 0 homology will be $\mathbb{Z}^2$, and all other homology groups will be trivial.


I hope this helps ^_^

HallaSurvivor
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