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1. • CommentRowNumber1.
   • CommentAuthorMurat_Aygen
   • CommentTimeFeb 18th 2018
   • (edited Feb 19th 2018)
   • PermaLink
   Author: Murat_Aygen
   Format: MarkdownItex**Theorem:** Let $K$ be a complex with $r$ combinatorial components. Then $H_{0}(K)$ is isomorphic to the direct sum of (not $r$ but) $2r$ copies of the group $Z$ of integers. **Proof for one component:** As 0-cycles are synonymous with 0-chains, they have $\alpha_{0}$ many dimensions-of-freedom for sure. How many the 0-boundaries of our component have? Since the incidence matrix is a $\alpha_{0}$ by $\alpha_{1}$ matrix (see my previous posting with ‘homology’ tag), they have at least $\alpha_{1} - \alpha_{0}$ many for sure. And 1 more for the rank deficiency! The proof follows from this arithmetic.

   Theorem:

   Let $K$ be a complex with $r$ combinatorial components. Then $H_{0}(K)$ is isomorphic to the direct sum of (not $r$ but) $2r$ copies of the group $Z$ of integers.

   Proof for one component:

   As 0-cycles are synonymous with 0-chains, they have $\alpha_{0}$ many dimensions-of-freedom for sure. How many the 0-boundaries of our component have? Since the incidence matrix is a $\alpha_{0}$ by $\alpha_{1}$ matrix (see my previous posting with ‘homology’ tag), they have at least $\alpha_{1} - \alpha_{0}$ many for sure. And 1 more for the rank deficiency! The proof follows from this arithmetic.