To solve this problem, we need to evaluate the relational schema \(\text{team(name, city, owner)}\) with the given functional dependencies and assess the decomposition into relations \(\text{t1(name, city)}\) and \(\text{t2(name, owner)}\).

Step 1: Evaluate the normalization of the relation team.

• The given functional dependencies are:
  • \(\text{name} \rightarrow \text{city}\)
  • \(\text{name} \rightarrow \text{owner}\)
• Since \(\text{name}\) is the only determinant for the attributes \(\text{city}\) and \(\text{owner}\), it implies that \(\text{name}\) is a candidate key for the relation.
• Every determinant is a candidate key, so the relation is in BCNF. However, since every relation in BCNF is also in 3NF, the relation is also in 3NF.

Step 2: Check the decomposed relations t1 and t2.

• Relation t1(name, city):
  • Functional dependency: \(\text{name} \rightarrow \text{city}\)
  • \(\text{name}\) is a candidate key for this relation. Therefore, it is in BCNF.
• Relation t2(name, owner):
  • Functional dependency: \(\text{name} \rightarrow \text{owner}\)
  • \(\text{name}\) is a candidate key for this relation. Therefore, it is in BCNF.

Step 3: Check for lossless join property.

• The original relation team can be reconstructed from the decomposed relations t1 and t2 using the common attribute \(\text{name}\) which is a key in both decompositions.
• Since the decomposition is based on projecting the attributes determined by \(\text{name}\), the join is lossless because it is a projection join.

Conclusion:

• The relations t1 and t2 are in BCNF.
• The decomposition is lossless because the projections on \(\text{name}\) allows the original table to be reconstructed without loss of data.
• The statement \(\text{"The relations t1 and t2 are in BCNF"}\) is true.
• The statement \(\text{"The decomposition constitutes a lossless join"}\) is also true.