Let $G = (V, E)$ be a simple connected eulerian graph, I was asked to prove that it cannot be bipartite if $(V, E - \{ e_1,e_2,e_3 \})$ is also eulerian.
While processing my proof I came up with this simple, connected and eulerian graph: $$v_1-v_2-v_3-v_4-v_5-v_6-v_1$$ $G$ is also bipartite e.g. let $A = \{v_1,v_3,v_5\}$,$B = \{v_2,v_4,v_6\}$. If I would remove the edges $v_1v_2,v_2v_3,v_3v_4$, $$v_4-v_5-v_6-v_1, v_2, v_3$$ is still eulerian, what am I missing?
Edit: I now noticed that an eulerian graph is defined as a graph that conatins an eulerian cycle my example only has an eulerian path therefore the graph is not eulerian.
Since $G$ is Eulerian, each vertex has even degree, and the sum of the degrees in each of its two parts is even. Since $G$ is bipartite, all of the edges are between the two parts, so removing $3$ of them reduces the sum of the degrees of the vertices in each part by $3$, leaving an odd total. This is possible only if at least one vertex in each part now has odd degree, in which case the new graph cannot be Eulerian. (In fact the only way to remove $3$ edges from an Eulerian graph and not produce a vertex of odd degree is to remove a triangle of edges, which does not exist in a bipartite graph.)