# Why is 2-connectedness characterized by the existence of an ear decomposition ?

Reading the book "Introduction to Graph Theory" (West, 2nd ed.) I have come across the following definition and statement: Why is this statement true ?

Suppose I have a graph $G$ that consist of a cycle $C$ and a path $P$ that share no vertices. Then $G$ has an ear decomposition ? Well, I can decompose $G$ into $P_o = C$ and $P_1 = P$ ? $P_o$ is a cycle and $P$ is an ear of $P_0 \cup P_1$ - $P$ is a maximal path whose interval vertices have degree $2$ in $G$ ? But $G$ is not connected and therefore is $0$-connected and not $2$-connected ? • Which book Introduction to Graph Theory? Amazon UK lists at least five books with that exact title (Chartrand and Zhang, Trudeau, Voloshin, West, Wilson). – David Richerby Feb 15 '14 at 22:05
• From the image I think it's the one by West. – Hsien-Chih Chang 張顯之 Feb 16 '14 at 0:03
• The one by West 2nd edition – Shuzheng Feb 16 '14 at 7:03

"Ear decomposition" is not a term I've heard before but it's incorrectly defined, as you've observed, since it doesn't require that the graph be connected, let-alone 2-connected. An "ear" needs to be defined as having each endpoint adjacent to at least one further vertex of $G$. (Actually, each endpoint would have to be adjacent to at least two further vertices, or the path wouldn't be maximal.)
• Yes, I'm sure: your counterexample is simple and clearly meets the definitions given in the image. I agree that it's surprising to see such an elementary mistake in a book, especially if the mistake was in the first edition, too. But, for example, see Diestel's book Graph Theory (online version): Whitney's theorem is Proposition 3.1.1; the definition of $H$-path (equivalent to an "ear") is at the bottom of page 7 in Section 1.3. The theorem is the same; the definition requires $H$-paths to meet the rest of the graph at both ends. – David Richerby Feb 16 '14 at 9:44