Why does if A is a spanning tree which doesn't have $e_1$ then $A\bigcup\{e_1\}$ has a unique cycle? I am studying the algorithm of Sollin and we recently studied a lemma:
Let be G a graph which values are diffferent on the edges.

We sort the edges  $e_1,e_2,...e_m$ such as $v(e_i)&lt;v(e_j)$
Every tree of minimal value has at least $e_1$ et $e_2$


Proof
let be A a spanning tree which doesn't have $e_1$
$A\bigcup\{e_1\}$ has a unique cycle
let be $e_i \in$ cycle, $e_i\ne e_1$ then $A\bigcup\{e_1\}-\{e_i\}$ is a tree which value is &lt; $v(A)$ because $v(e_i)&gt;v(e_1)$

I don't understand why would $A\bigcup\{e_1\}$ would have a unique cycle?
Do you have a justification? A graphical example?
In a tree, there is exactly one path between any pair of vertices. When you add edge e1 you close that unique path to form a unique cycle.