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Example text

T*1 . extltsl x, si x,). e. e. 0 ) Ue I * f*1. _ (s; I, s;gl s; I, ). 11. so let s[ 1 +o ). )be only r +O X an extension to verify that of the X by X,. k_th component of '\a,r; s l r ' ) * . t is exact. Let x = (xi, jpi), I'= Then the we get following conmutative ,^+ o (xi , jei ) and E = (Errrtr). diagrarn o U I + o I (s; I)k - (s;I')k e O -'O xi r acl I U. t o- xi I I and the third first to see middle that exact. ,n of sequence dimensions involved Extl(six,si x'), map. s": L(M,0)-L( functors.

2 . 1 ) H o m ( z , c ' - I ) ! u o m ( ! 1, c - ^ - ' x ) * + L +l Hom(z,c*^t' I). 3)z is injective, hence also being injective c*k*lx must be itserf im a < c*k*1 I, "o indecomposabl-e. 3). so{ lc-qr, I q t x} is a infinite set of pair\tise non iso- I nom(c-9p,'x) morphic modules. 1' for q: k. 'lO Proposltlon: LetX€ a) Then the following X is preprojectlve. li) There is a g>O indecomposable ! the are equivalent: C+q X = O. such that Horn (Y, X)+ O only Dually ). statements i) ili) b) l , l, 0 [( for (M, € t following a finite number of n ).

R . sn Q). rr source. s. s|*f It) = o and rt Let X€ _ I a)X=CC'xeP,whereP€ L ( is injective. M , 0 ) . Then L (M,Q)isprojective. r) s ; + 1 E r ) r=* r " n - f 1 . : = . ; ; - . ; - " . ; ; - = = ( s k ( d t usn; . . e. e. C- X = O, or I = C* C Ext' Proof: a) ( c b) If If k €f = a sink, t* (dirn X ) . F* as bimodule. enough to prove a). 4). 3), and. 6) is is sequence with k = 1. 3) and F1 vrould. be injective this is impossible, >1 by general assurnptlon. , = 1, s; F1))1 = d\, because the a vector and lll El = O by easy to see, that a sink ( dtun (s;...