$G$ is a connected graph. Show that if $A,B$ are bonds, and $e \in A \cap B$, there is a bond $C \subseteq A \cup B \setminus \{e\}$ $G$ is a connected graph. Show that if $A,B$ are bonds, and $e \in A \cap B$, there is a bond $C \subseteq A \cup B \setminus \{e\}$
A bond is a set of edges $X$ so that $G \setminus X$ has two connectivity components.
Any hints and assistance would be much appreciated!  
 A: Let $G = (V,E)$ be a connected graph.
Every bond of $G$ is a cut. 
A cut $F \subseteq E$ is a set of edges of $G$ such that there are vertex sets $V_1, V_2 \subseteq V$ with $F = E(V_1,V_2) := \{vw \in E:: v \in V_1 \wedge w \in V_2\}$. 
The set of all cuts (including the empty cut) of a graph $G$ is the cut space $\mathcal C^*(G)$ of $G$.
The cut space $\mathcal C^*(G)$ is a vector space over $GF(2)$, 
i.e., the field with only two elements $\{0,1\}$, 
with the symmetric difference $\dotplus$ of sets as addition (in detail $A \dotplus B := (A\cup B)\setminus (A \cap B)$ for sets $A$ and $B$), 
that is, two cuts $F_1,F_2 \in \mathcal C^*(G)$ can be added $F_1 \dotplus F_2 \in \mathcal C^*(G)$.
Back to your question: If $A,B \in \mathcal C^*(G)$ are bonds, 
then $A\dotplus B \in \mathcal C^*(G)$ is a cut. 
Note that $e \notin A \dotplus B$ because $e \in A\cap B$. 
A bond is also a minimal non-empty cut. 
Note that in a connected graph no cut is empty. 
Since $A \dotplus B$ is a cut, 
there must be a minimal cut contained in it, 
i.e., there must be a bond $C \in \mathcal C^*(G)$ 
with $C \subseteq A \dotplus B \subseteq (A \cup B)\setminus \{e\}$.
