Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Functional programming, design and analysis of algorithms / 8 / b

True. A tree has exactly one path between any two vertices. For $e\notin T$, that path between the endpoints of $e$, together with $e$, is a fundamental cycle. Every cycle in $T\cup \{e\}$ must contain $e$, since $T$ itself is acyclic; deleting $e$ from such a cycle leaves that same unique path. Therefore the cycle is unique. Minimality of the spanning tree's weight is not needed for this assertion.

True. Delete $e\in T$; the resulting two connected components define a cut of a graph. If a crossing edge $f$ had $w(f)<w(e)$, the exchange $T-e+f$ would be a spanning tree of smaller total weight, contradicting the minimum spanning tree assumption. Thus $e$ is a light edge crossing a cut for this fundamental cut. It need not be uniquely light.