[CLRS 23.1-6] Unique Light edges lead to a unique spanning tree

23.1-6 [CLRS]:Show that a graph has a unique minimum spanning tree if, for every cut of the
graph, there is a unique light edge crossing the cut. Show that the converse is not
true by giving a counterexample.

Here, let's use the theorem 23-1, according to which we can select $A$ to be a subset of $E$, the set of edges that is included in some minimum spanning tree. Then,
Hypothesis-1 : We can always make $A$ such that it can respect any cut $(S,V-S)$.
Eventually, that is very easy to do. We can just ignore all the edges that lie on the cut (that is, connect the two components produced by the cut), and put the rest of the edges into $A$. This leads to an $A$ that respects the underlying cut.

Now, according to the question, each cut corresponds to a light edge. Now we know that we have a total of $2^{n-1}-1$ possible cuts for $n$ vertices. So, each of these cuts, points towards an edge, that, according to the theorem is safe to add into $A$. So we add all the corresponding edges of all the possible cuts.

Hypothesis-2: $|A|=|V|-1$, when $A$ is made by adding all the corresponding light edges of every possible cut.
Let's assume that we'll have $|A|=|V|$. Since there are $|V|$ vertices in total, we're sure to have a cycle in $G_A$. Now, when we make cuts such that we have one vertex of the cycle in one cut, and all the other vertices in the other one, then we'll always leave one edge of the cycle out. Thus, contradicting the assumption. Therefore the hypothesis is true.

Now, we know that $A$ has exactly $|V|-1$ edges in it. Moreover, $A$ will always come to be the same set, on the same graph $G$.

And, we know that a set of $|V|-1$ different edges, for $|V|$ vertices, always leads to an MST.

CONVERSE:
The converse states that, "If a graph has a unique MST, then for every cut of the graph, there is a unique light edge crossing the cut."

[Photo credits : https://github.com/gzc/CLRS/blob/master/C23-Minimum-Spanning-Trees/23.1.md]