6

I'm reading Chartrand's A First Course in Graph Theory and there's this problem I'm struggling with. Would appreciate some guidance on the proof strategy.

The problems reads as follows:

Let $G$ be a connected graph and let $e_1$ and $e_2$ be two edges of $G$. Prove that $G - e_1 - e_2$ has three components if and only if both $e_1$ and $e_2$ are bridges in $G$.

For the forward direction ($\Rightarrow$): If $G - e_1 - e_2$ has three components, I need to show both edges must be bridges. I'm thinking that if either edge were not a bridge, then removing it wouldn't disconnect the graph, which would contradict having three components after removing both edges.

For the backward direction ($\Leftarrow$): If both $e_1$ and $e_2$ are bridges in $G$, I need to show that $G - e_1 - e_2$ has exactly three components. Since each bridge increases the number of components by 1 when removed, removing two bridges should give us three components total. However, I'm not sure how to formalize this argument rigorously, especially regarding whether the bridges share a common vertex or not.

Any hints or suggestions for how to structure this proof would be greatly appreciated!

RobPratt
  • 52,643
Jisbon
  • 131

2 Answers2

5

We will briefly show here that if $e_1$ and $e_2$ are bridges in $G$, then $G - e_1 - e_2$ has exactly three components.

Since $ e_1 $ is a bridge in the connected graph $G$, the graph $ G- e_1 $ consists of two components. Denote by $ H $ the component containing $ e_2 $, and by $ F $ the other component. Because $ e_2 $ is a bridge in $ G $, it is also a bridge in $ H $ (otherwise $ H - e_2 $ would be connected, and the graph $ (H - e_2) + e_1 + F =G-e_2 $ would also be connected). This essentially completes the argument. The graph $ H - e_2 $ has two components, plus the additional component $ F $.

kabenyuk
  • 13,900
  • 1
    It would be nice to put explicitly here that you are proving the direction $e_2,e_1$ both bridges $\implies$ $G\setminus {e_2,e_1}$ has exactly $3$ components. – Mike Oct 05 '25 at 15:35
  • 1
    @Mike thank you—you're absolutely right. I just looked at my answer with fresh eyes and realized that pointing this out is utterly essential. – kabenyuk Oct 06 '25 at 12:50
1

A quick argument for the forward direction: removing one edge in an arbitrary graph can increase the number of components by at most $1$. (Show the contrapositive: adding an edge to an arbitrary graph can only decrease the number of components by $1$. This is true because any edge will connect at most two components and turn them into one.) Now, if $G-e_1-e_2 (=G-e_2-e_1)$ has three components then by the above we know that both $G-e_1$ and $G-e_2$ have at least two components. Definitionally, this makes them both bridges.

Steven Stadnicki
  • 53,503