A statement in my course says that : “A connected graph $G$ contains an Euler $uv$-path if and only if all vertices except possibly $u,v$ are even.”
I agree with the $\implies$ direction, but in the other way, I do not. If we consider a cycle with $\geq$ 3 vertices then if $u$ and $v$ are distinct, there will be no $uv$-path linking $u$ and $v$ that does not repeat an edge twice and covers all edges. In fact in a cycle there are only two paths linking $u$ and $v$ and none of them seems to be Eulerian to me, even though all vertices in a cycle (that is connected) are even. The proof I was given treats the case where $uv$ belongs to the path or not. So this means that the statement wanted to say $\forall u,v$ and that is the mistake I think. Wouldn’t it be true if it was $\exists u,v$ ? So that in a cycle there exists an Eulerian path (itself).
What do you think ?