it says here that Z is a conflict, can someone explain why, please?

(a) Transaction read_item(X); write_item(X); read_item(Y); write_item(Y); Transaction T2 read_item(Z); read_item(Y); write_item(Y); read_item(X); write_item(X); Transaction Tz read_item(Y); read_item(Z); write_item(Y); write_item(Z); (b) Transaction T Transaction T2 Transaction T3 read_item(Z); read_item(Y); write_item(Y); read_item(Y); read_item(Z); Time read_item(X); write_item(X); write_item(Y); write_item(Z); read_item(X); read_item(Y); write_item(Y); write_item(X); Schedule E (c) Transaction T Transaction T2 Transaction Tz read_item(Y); read_item(Z); read_item(X); write_item(X); write_item(Y); write_item(Z); Time read_item(Z); read_item(Y); write_item(Y); Figure 21.8 Another example of serializability testing. (a) The read and write operations of three transactions T1, T2 and T3. (b) Schedule E. (C) Schedule F. read_item(Y); write_item(Y); read_item(X); write_item(X); Schedule F (d) Y Equivalent serial schedules (T. T None Reason Y Y, Z Cycle X(T Cycle X(TA T2), Y(T2 T) T2),YZ (T2 T3),Y(Tz T) T3 (e) X,Y Equivalent serial schedules TA T Tz TT2 Y Y, Z T3 (f) Equivalent serial schedules T) T. T3 T, Tz Tz T2T T3 Figure 21.8 (continued) Another example of serializability testing. (d) Precedence graph for schedule E. (e) Precedence graph for schedule F. (f) Precedence graph with two equivalent serial schedules