A plane graph $G$ is $2-$face colorable if and only if $G$ is eulerian - counter example

Question

Actually, proof is asked in here: Planar graph has an euler cycle iff its faces can be colored with 2 colors. But my question is not about proof but about why is the following graph not a counter example:

I read the definition of eulerian graph in the book and looked it up in Wikipedia but both says eulerian graph is a graph that contains eulerian cycle (or tour) or a graph with every vertex of even degree. So, above graph is a plane graph and 2-face colorable, but it does not contain eulerian cycle. As I saw in here, above graph is semi-eulerian as it contains an eulerian path but not an eulerian cycle. So why this graph is not a counter example for the argument in the title? Thanks in advance.

Answer 1

If you look at the dual graph of your example, (to see how to color the faces) you see that you get a loop edge because of the pendant vertex in your original graph.

By definition, (at least from my book and other places such as here )a graph must not contain any loop edges in order to be colorable.

It is even mentioned on the Wikipedia page for vertex coloring:

"...Since a vertex with a loop (i.e. a connection directly back to itself) could never be properly colored, it is understood that graphs in this context are loopless."

Answer 2

The theorem as you stated it isn’t true, as your example shows. I think the correct theorem is “A map $G$ is $2$-face colorable if and only if $G$ is Eulerian.” A map (not to be confused with a “map graph”) is a $3$-connected planar graph. It’s possible the theorem can be corrected in a different way be redefining what a face-coloring is, but I’m more comfortable saying that your example is not a map than saying that your example is not face-colorable.