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Definition of Betti number at http://en.wikipedia.org/wiki/Betti_number

The $n^{th}$ Betti number represents the rank of the $n^{th}$ homology group, denoted $H_n$ "which tells us the maximum amount of cuts that can be made before separating a surface into two pieces"

Is it true for all $n$ or just for $n=1$? What does it mean by "separating a surface into two pieces"? Is it related to connected components?

Najib Idrissi
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Vrouvrou
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  • I stil dont inderstand i know that $H_1=\ker(\partial: C_1\rightarrow C_0)/\text{Im}(\partial: C_2\rightarrow C_1)$ ans $\beta_1=\dim H_1$ so what it means ? – Vrouvrou Jul 28 '14 at 09:13

1 Answers1

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"Cutting a surface into two pieces" corresponds to the case of a $2$-manifold (= surface) and $n=1$, in more generality it's the rest of the sentence: "two pieces or 0-cycles, 1-cycles, etc."

In the case of a surface and $n=1$, the first Betti number $b_1$ does correspond to how many "cuts" you can make before getting more than one connected components.

  • With a sphere, $b_1 = 0$ and you can't make any cuts (in the form of a closed curve, basically a circle) without separating the sphere into two connected components.
  • With a torus, $b_1 = 2$ and you can make two cuts (in the form of a circle) and still keep only one connected component (try to visualize it). If you make three, you'll get at least two components.

Keep in mind that this is only an informal explanation, and with increasing $n$ it's harder and harder to understand what $b_n$ "means" (how many times you can cut your space with an $n$-cycle and still keep one $(n-1)$-"piece").

Najib Idrissi
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  • Thank you, please What is the book where i can fined all this ? – Vrouvrou Jul 28 '14 at 09:27
  • Hatcher's Algebraic Topology is commonly cited and is available for free, you can also look at this thread. – Najib Idrissi Jul 28 '14 at 09:30
  • I have Hatcher's book but i don't fined this – Vrouvrou Jul 28 '14 at 09:34
  • I don't know if he explicitly explains what Betti numbers represent, but he does explain what the homology groups represent, and understanding of $b_n$ should come naturally afterwards. – Najib Idrissi Jul 28 '14 at 09:37
  • Najib, is the statement $b_1(\mathbb{R}^2)=0$ equivalent to the Jordan curve theorem? – goblin GONE Apr 07 '17 at 01:28
  • @goblin No, it's much weaker than that. $b_1(\mathbb{R}^2) = 0$ means that the rank of the abelianized of $\pi_1(\mathbb{R}^2)$ is zero, i.e. all elements in the abelianized are torsion. Even the statement "$\pi_1(\mathbb{R}^2) = 0$" is much weaker: it says that given any continuous map $S^1 \to \mathbb{R}^2$, you can extend it to the disk $D^2$. The Jordan curve theorem says, very roughly speaking, that given an injective map $S^1 \to \mathbb{R}^2$, you can extend it to an injective map $D^2 \to \mathbb{R}^2$. – Najib Idrissi Apr 07 '17 at 07:12