How to prove that if G has no odd cycles then the chromatic number of G is 2

Question

$G$ has no odd cycles $\iff$ the $X(G)=2$

I already proved the reciprocal but I'm having trouble with this.

Here is my thinking:

Let be $G=(V,E)$. If $G$ has no cycles, then $G$ is a tree so $X(G)=2$ since trees are bipartite.

If $G$ has one cycle, I know it's an even cycle. Let be $C_1=\{v_1,...,v_n\}$ an even cycle of $G$. We can define $k:C_1\rightarrow \{a,b\}$ a coloring of c.

$k(v_1)=a, k(v_2)=b, ... , k(v_n)=b$

Since $k=a$ for odd indexes, $k=b$ for even indexes, and $n$ is even. So $X(C_1)=2$.

But I don't know how affirm that $X(G)=2$. Not to mention $G$ with more than one cycle.

Any help will be appreciated, thanks in advance.

Possible duplicate of Proof a graph is bipartite if and only if it contains no odd cycles (Bipartite implies $X(G)\leq 2$ because you can color the left side red and the right side blue.) Note that a graph with no edges (like a completely disconnected graph or a graph with a single vertex) will have chromatic number $1$. — Arthur, Nov 25 '19 at 18:33

score 0 · Accepted Answer · answered Nov 25 '19 at 18:37

0

Recall that a graph $G$ is bipartite if and only if it $\chi(G) = 2$. Also, recall that a graph is bipartite if and only if it has no odd cycles.

These two statements combined give our desired result.

answered Nov 25 '19 at 18:37

Ekesh Kumar

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How to prove that if G has no odd cycles then the chromatic number of G is 2

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