Eulerian orientations on connected countably infinite graph

Question

Let $G=(V,E)$ be a locally finite graph with $V$ countably infinite. Assume that each vertex has even degree. Does there necessarily exist an Eulerian orientation on $G$? That is, is there a way to direct the edges of $G$ so that the number of edges going into a vertex $v$ is same as the number of edges going out of $v$? I know that the answer is yes in the case of $2$-connected graphs as is mentioned here

2-connected graph has a strongly connected orientation

However, what if we drop the $2$-connectedness condition? That is, if we drop the property that $G$ is still connected upon the deletion of any one edge can we still find an Eulerian orientation?

Answer 1

 

    "...can we [...] find an Eulerian orientation?" -- YES

A slightly more general result holds, as below (see the last Remark at the bottom of this note).

                ASSUMPTION (once and forever)

Let $\,G=(V\,E)\,$ be a locally finite graph such that every vertex has even degree.

  THEOREM 1

For every $\,e\in E\,$ there exists a positive $\,n\in\Bbb Z\,$ and an injective map $\,\pi: \Bbb Z/n\to E\,$ such that $\,\pi(0)=e\,$ and

$$ \forall_{k\in\Bbb Z/n}\quad \pi(k)\,\subseteq\,\pi(k-1)\cup\pi(k+1) $$

In other words, $\,\pi\,$ is a finite cycle when $\,n>1,\,$ and $\,\pi\,$ is a path that is infinite in both directions when $\,n=1\,$ -- then $\,\Bbb Z_1=\Bbb Z.$

DEFINITION A path $\,\pi\,$ as above is called an orbit.  (By this definition, every orbit is non-empty.)

    THEOREM 2

Let $\,H=(W\,F)\,$ be a graph such that $\,W\subseteq V\,$ and $\,F\subseteq E,\,$ and every vertex $\,w\in W\,$ has even degree in $\,H.\,$ Furthermore, let $\,D\,:=\,E\setminus F\ne\emptyset\,$ and $\,U:=\bigcup D\,$ (so that $\,U\subseteq V).\,$ Then there exists an orbit $\,\pi\,$ in graph $\,(U\,D).$

Proof (of Theorem 2)

It's clear that every vertex $\,u\in U\,$ has even degree in graph $\,(U\,D)\,$ (and $\,D\ne\emptyset).\,$ Thus, Theorem 2 follows from Theorem 1.

    THEOREM 3

Graph $\,G\,$ admits an Eulerian orientation.

Proof (of Theorem 3)

By straightforward transcendental mathematical induction or by Kuratowski-Zorn Theorem (also called Zorn Lemma), $\,E\,$ is a union of pairwise disjoint orbits. Select either one of the two orientations for every orbit. This proves our Theorem 3.   Great!

REMARK   Theorem 3  holds for arbitrary cardinalities of $\,\,V$ and $\,E.\,$ (There is no need to assume that $\,V\,$ nor $\,E\,$ are infinite or countable).  Furthermore, if we assumed that every infinite cardinal number is even then we do not need to say anything about graph $\,B\,$ being locally finite.

EDIT:   @bof has wondered at one time about the non-local case. It takes only one extra consideration.

Let $\,S(v):=\{w\in V: {v\,w}\in E\subseteq E.\ $ Since every vertex $\,v\in E\,$ has even degree (finite or not) it means that there is a fixed-point free involution $\,i_v:S(v)\to S(v)\,$ for each $\,v\in V.\,$

Each time we start a new orbit by choosing $\,\pi(0)\in D\,$ we also choose the orientation $\,(v^-\,v^+)\,$ of $\,\pi(0)=\{u\,w\},\,$ i.e. $\,\{v^-\,v^+\}\,=\,\{u\,w\}.\,$ Now the whole orbit is determined uniquely by recursively defining new $\,\pi(n+1):=i_v(\pi(n))\,$ for $\,v\in\pi(n)\,$ and $\,n>0\,$ but $\,v:=v^+\,$ for $\,n=0;\,$ and new $\,\pi(n-1):=i_v(\pi(n))\,$ for $\,v\in\pi(n)\,$ and $\,n<0\,$ but $\,v:=v^-\,$ for $\,n=0;\,$.   Great!