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In my book, it's written that the following are equivalent. I don't understand why it is so.

  • G is bipartite;
  • G is 2-colourable;
  • G does not contain any odd length cycles.
William
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Gigili
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  • They're equivalent in that each implies both of the others, in case your question isn't about the truth of the statement, but of terminology. That is, if I'm about to start proving a theorem about graph coloration, I probably wouldn't start by saying "Suppose G has some odd-length cycle", but instead "Suppose G is not 2-colorable", even though they mean the same thing. – Eric Tressler May 25 '13 at 06:58

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If $G$ is bipartite with parts $V$ and $W$, color the vertices in $V$ red and the vertices in $W$ blue. Conversely, if $G$ is $2$-colorable, splitting the vertices by color gives a bipartition. This shows that the first two are equivalent.

If $G$ is bipartite with parts $V$ and $W$, any path must alternate between vertices in $V$ and vertices in $W$. Suppose that you start at a vertex in $V$; then after an odd number of steps you must be in $W$, and after an even number of steps you must be back in $V$. In particular, to return to your starting point requires an even number of steps, and therefore $G$ has no odd cycles. For the converse see this question and its answer, or this answer.

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Brian M. Scott
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