A question was put up: let $G=(V,E)$ be a connected, undirected graph with weights on its edges. Let $T_1 = (V,E_1)$ and $T_2 = (V,E_2)$ be two different minimal spanning trees of $G$. Prove that in the United Graph $(V, E_1\cup E_2)$ there is a cycle with at least one pair of edges with the same weight
I am trying to understand what may be wrong with the following (simple) proof:
Take an edge $e$ that is in $T_1$ and not in $T_2$. Add it to $T_2$ and now $T_2+e$ has a cycle. Now it is not possible for $e$ to be the heaviest (greater than all other edges) in that cycle because otherwise, it wouldn't have appeared in $T_1$ (Kruskal). It is also not possible for the largest edge in that circle which is not $e$, let's call it $e_{\max}$, to make $w(e_\max) > w(e)$ because then $e_\max$ wouldn't have appeared in $T_2$. Therefore edge $e$ has the same weight as the maximal-weight edge in that cycle.