Let $G$ be an undirected connected graph in which every edge has a positive integer weight. Suppose that every spanning tree in $G$ has even weight. Which of the following statements is/are TRUE for every such graph $G$ ?

Let $T$ be any spanning tree of $G$. We are given that the weight of every spanning tree in $G$ is even.

1. Edge Exchange Property: For any edge $e = (u, v) \notin T$, adding $e$ to $T$ forms a unique fundamental cycle $C_e$ containing $e$ and a path $P \subseteq T$ connecting $u$ and $v$. For any edge $e^{\prime} \in P$, the set $T^{\prime} = (T \cup \{e\}) \setminus \{e^{\prime}\}$ is also a valid spanning tree of $G$.

   The weight of $T^{\prime}$ is:

   $$w(T^{\prime}) = w(T) + w(e) - w(e^{\prime})$$

   Since both $w(T)$ and $w(T^{\prime})$ are even integers, their difference must be even:

   $$w(e) - w(e^{\prime}) \equiv 0 \pmod 2 \implies w(e) \equiv w(e^{\prime}) \pmod 2$$

2. Parity in Cycles: This implies that every edge in the cycle $C_e$ must have the same parity (either all even or all odd). Since any cycle $C$ in $G$ can be decomposed into fundamental cycles (or can have a tree containing all but one of its edges), all edges in any cycle $C$ must share the exact same parity. Therefore, statement (D) is TRUE.

3. Counterexample for (A), (B), and (C): Consider a graph $G$ that is itself a tree (which has no cycles). Since $G$ has only one spanning tree (itself), as long as the sum of all edge weights is even, the condition holds. For example, a tree with two edges of weight $1$ (odd) and one edge of weight $2$ (even) has total weight $4$ (even). In this graph, not all edges are even, nor are all edges odd, disproving (A) and (B). Moreover, if a graph consists of two disjoint cycles connected by a bridge, one cycle could have all even edges and the other all odd edges, disproving (C).

Hence, only statement (D) is correct.