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  "body": "\n## Proof\n\nWe prove the easy direction by alternation of sides, and the converse by colouring with the parity of the distance.\n\n1. ($\\Rightarrow$) Let $(A, B)$ be a bipartition of $G$, as in the definition of a [[bipartite|07DVHJ8N0A]] graph. Let $(w_0, \\dots, w_k)$ be a [[cycle|07CT9RDY7Q]], and say $w_0 \\in A$; the case $w_0 \\in B$ is symmetric.\n2. Each edge has one end in $A$ and one in $B$. By induction on $t$, $w_t \\in A$ for even $t$ and $w_t \\in B$ for odd $t$.\n3. Since $w_k = w_0 \\in A$, the length $k$ is even. So $G$ has no cycle of odd length.\n4. ($\\Leftarrow$) Lemma: if $G$ has a closed [[walk|0VJZM0F9YM]] of odd length, then $G$ has a cycle of odd length. We use [[strong induction|0KMVFSHFGZ]] on the length $k$.\n5. Let $W = (w_0, \\dots, w_k)$ with $w_k = w_0$ and $k$ odd. Since no vertex is adjacent to itself, $k \\ne 1$. So $k \\ge 3$.\n6. Case 1: $w_1, \\dots, w_k$ are pairwise distinct. The edges are $\\set{w_{t-1}, w_t}$ with indices modulo $k$. Two of them, for $s \\ne t$, could only agree if $w_{s-1} = w_t$ and $w_s = w_{t-1}$.\n7. That would give $t \\equiv s - 1$ and $s \\equiv t - 1$ modulo $k$, so $k \\mid 2$, which is false. So the edges are distinct, $W$ is a [[circuit|0PD57PSHNR]], and it is a cycle of odd length.\n8. Case 2: $w_i = w_j$ for some $1 \\le i \u003c j \\le k$. Then $W_1 = (w_i, w_{i+1}, \\dots, w_j)$ is a closed walk of length $j - i$.\n9. Also $W_2 = (w_j, w_{j+1}, \\dots, w_k, w_1, \\dots, w_i)$ is a closed walk of length $k - (j - i)$, using $w_k = w_0 \\sim w_1$.\n10. Both lengths are between $1$ and $k - 1$, and they add up to the odd number $k$. So one of them is odd, and the induction hypothesis gives an odd cycle. This proves the lemma.\n11. Now assume that $G$ has no cycle of odd length. By the [[axiom of choice|0HTGB13FE2]], choose a vertex $r_C$ in each [[connected component|01VVP10MWJ]] $C$ of $G$. For finite $G$ no choice principle is needed.\n12. For a vertex $v$ in the component $C$, the vertex $v$ is [[reachable|0J8XE7W68F]] from $r_C$. So the [[distance|0WPZ5YPAGR]] $\\delta(v) = d(r_C, v)$ is a natural number.\n13. Let $A = \\set{v : \\delta(v) \\text{ even}}$ and $B = \\set{v : \\delta(v) \\text{ odd}}$. They are disjoint, and their union is the vertex set.\n14. Let $xy$ be an edge. Then $x$ and $y$ are reachable from each other, so they lie in the same component $C$. Let $r = r_C$.\n15. Suppose $\\delta(x)$ and $\\delta(y)$ have the same parity. Take a walk $P$ from $r$ to $x$ of length $\\delta(x)$, and a walk $Q$ from $r$ to $y$ of length $\\delta(y)$.\n16. Follow $P$ from $r$ to $x$, then the edge $xy$, then $Q$ backwards from $y$ to $r$. This is a closed walk of length $\\delta(x) + 1 + \\delta(y)$, which is odd.\n17. By the lemma, $G$ has a cycle of odd length, a contradiction. So $\\delta(x)$ and $\\delta(y)$ have different parities.\n18. Hence every edge has one end in $A$ and one end in $B$, and $G$ is bipartite.\n\n## Idea\n\nAlong any walk the side changes at each step, so a closed walk in a bipartite graph has even length. Conversely, colour each vertex by the parity of its distance from a root. An edge inside one colour class would close an odd walk, and an odd closed walk always contains an odd cycle.\n",
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