# Bridge divides Graph into Two Components

## Theorem

Let $G$ be a connected graph.

Let $e$ be a bridge of $G$.

Then the edge deletion $G - e$ contains exactly $2$ components.

## Proof

Let $G$ be a connected graph and $e = u v$ be a bridge of $G$.

By definition of bridge, $G - e$ has to be of at least $2$ components.

Aiming for a contradiction, suppose $G - e$ were of more than $2$ components.

Let $G_1, G_2, G_3$ be $3$ of those components such that $u \in G_1$ and $v \in G_2$.

Note that $u$ and $v$ cannot both be in the same component or $e$ would not be a bridge.

Now let $e$ be replaced so as to connect $G_1$ and $G_2$ again.

Then $G_3$ would still be disconnected from the rest of $G$, and $G$ itself would therefore be a disconnected graph.

From this contradiction it follows that $G - e$ can only be of two components.

$\blacksquare$

## Sources

- 1977: Gary Chartrand:
*Introductory Graph Theory*... (previous) ... (next): $\S 2.4$: Cut-Vertices and Bridges