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Let $G$ be a graph. I want to show that $E(G)$ is disjoint union $C\cup D$ where $C$ and $D$ belong to cycle and cut space respectively.

hhh
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user108209
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1 Answers1

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Note that the result says that we can partition $G$ into $V_1,V_2$ in such a way that all vertices in $G[V_1]$ and $G[V_2]$ have even degree. A proof of this is given in Combinatiorial Problems and Exercises (Lovász):

If all the vertices have even degree then we're done. Assume that $v$ is a vertex of odd degree with neighbours $N$. Let $G'$ be the graph given by removing $v$ and then adding an edge modulo $2$ between all pairs of elements of $N$.

By the induction hypothesis we can partition $G'$ into $W_1,W_2$.

Without loss of generality we can assume that $|N \cap W_1|$ is even and $|N \cap W_2|$ is odd, so we can set $V_1=W_1 \cup \{v\}$ and $V_2=W_2$ and check that everything has even degree as required.

Raoul
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  • I was wondering if you could give me the number of page for your claim in Lovasz's book – user108209 Oct 24 '14 at 07:42
  • It's (part of) question 5.17 (page 42). The answer is given on page 287. (There's also a hint on page 123.) – Raoul Oct 24 '14 at 19:37
  • It is clear that if degrees of all vertices of $C$ are even, then $C$ belongs to cycle space. But why is it true for cut space? – user108209 Oct 25 '14 at 14:11
  • As you note, the set $C$ of edges within $V_1$ and $V_2$ is in the cycle space (because all the vertices are even). The set $D$ of edges between $V_1$ and $V_2$ is a cut-set by definition. – Raoul Oct 25 '14 at 15:22