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Additional info for Indecomposable Representations of Graphs and Algebras
T(Y~and T(Y) would belong to R(t). This implies that the functor T: C --+ H(IJR, n) is dense and therefore T is an equivalence. Now, we are going to consider the individual extended Dynkin diagrams. In all cases, condition (i) is satisfied trivially and, in most cases, it is also very easy to see that the condition (iii)' is satisfied: one uses the properties of a simple regular representation X with T/(X) = 0 which are listed in the last column of the tables. These properties are satisfied for every simple homogeneous object, and therefore for every homogeneous representation at all.
FFF. Finally, 'P is determined by A 2 . Second, T(R) contains Eo, because (10 1) x 0 belongs to A 2 n B 1 in this case. On the other hand, T(R') contains a copy of E'l' because the kernel of the multiplication map- ~ FF is just N F' so that Al n 8 1 =1= O. Q) defines an equivalence of categories ping F G 0 GFF <321 . between the category LdG(F/G)G) of all representations 'P: UG 0 G(F/G)G ~ V G with a surjective 'P, and the category C of all representations (A 1 c.... A 2 c... Q) which 37 REPRESENTATIONS OF GRAPHS AND ALGEBRAS have the property that (A 1 c..........
Consequently, X n Y = 0, and E ~ Y c..... V - - E is the identity of E, i. e. the sequence splits. 3, Extl(E r , X) = Extl(E, C-rX) and the statement (2) follows. (4) To prove the last statement, consider the extension 0-----+ E I -----+ V -----+ C-(r-I)X -----+ 0, < * * E and C-(r- I)X ~ E I . Then 0 and E I is a quotient of V. 5. 4. Let (1JJl, S"l) be a realization of an extended Dynkin diagram (r, d) and C+ be the corresponding Coxeter functor. Let i be a source with respect to S"l. Let E be a representation of (WI, S"l) such that C+E =1= 0 and E i = O.
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