Let $T$ be a spanning tree. Prove that for cycles $C,D$, $E(C)\backslash E(T) = E(D)\backslash E(T)\implies C=D$ Let $T$ be a spanning tree of $G$. Prove that if $C$ and $D$ are cycles in G and $E(C)\backslash E(T) = E(D)\backslash E(T)$ then $C=D$.
So far I have that if $e$ is an edge in $E(C)$ then either $e$ is in $T$ or not. If $e$ it in $T$ then $e$ is in $E(C)\backslash E(T)=E(D)\backslash E(T)$ so $e$ is in $D$. Now suppose $e$ is in $T$.... and I'm stuck.
 A: Here's a proposition : 
If what you meant is that if $e$ is in $E(C)$ but not in $E(T)$, then $e$ is in $E(D)$ (and also the other way around), you got that right.
This means that any edge in one of $E(C)$ or $E(D)$, but not in both has to be in $E(T)$.
This in turn means that if $C$ and $D$ are edge-disjoint cycles, then $T$ contains all edges of $C$ and $T$ has a cycle, a contradiction.
So let $u_1u_2$ be some edge both in $E(C)$ and $E(D)$, and suppose $C \neq D$.
Write $C$ as $c_1c_2c_3 \ldots c_ic_1$ and $D$ as $d_1d_2 \ldots d_jd_1$, 
with $c_1 = d_1 = u_1$ and $c_2 = d_2 = u_2$.
Follow the path $c_1c_2c_3 \ldots c_k$, where $c_k$ is the first vertex such that $c_{k + 1} \neq d_{k + 1}$.  Since $C$ and $D$ share an edge, there must be some $h, h' > k + 1$ such that $d_h = c_{h'}$.  Essentially, $D$ leaves the $C$ cycle at point $k$ and reenters at point $h'$.
(If this is homework, you'd still have to explain why $k$ and $h'$ must exist)
But it turns out that $c_kd_{k + 1} \ldots c_{h'} c_{h' - 1} \ldots c_{k + 1}c_k$ is a cycle $B$, and that all of its edges are in exactly one of $C$ or $D$ (and not in both).
So all edges of $B$ are in $T$, and hence $T$ has a cycle.
