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By Sjoerd Beentjes

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Let [M ], [N ] ∈ Iso(A), then we see that G∗ ([M ] · [N ]) = M, N A m R FM,N [G (R)], [R]∈Iso(A) whilst G∗ ([M ]) · G∗ ([N ]) = G (M ), G (N ) B m FGS(M ),G (N ) [S]. [S]∈Iso(B) Therefore these two expressions are equal if G preserves the Euler-form and if it preserves the R sets FM,N for all objects M, N, R of A. Equivalently, this is so if G is an exact functor and ∼ if it is extremely faithful: it induces isomorphisms ExtiA (M, N ) −→ ExtiB (G (M ), G (N )) for all M, N ∈ Ob(A), for all i 0.

20. Let Q be a quiver. The dimension of the space Ext1Q (S(i), S(j)) equals the number cij of arrows in Q from i to j. 1 Generalities on quivers Proof. 14. When i = j, the domain of the map c ≡ cS(i),S(j) : Homk (S(i)s(α) , S(j)t(α) ) ∼ = Homk (S(i)l , S(j)l ) −→ l∈Q0 Homk (k, k) α:i→j α∈Q1 is the zero-dimensional vector space. Therefore, the cokernel of c equals the codomain of c, which in turn equals Ext1Q (S(i), S(j)). The isomorphism in the above equation shows that the dimension of the codomain of c is precisely cij .

Choose a vertex i0 such that Vi0 = 0, and take a non-zero vector v ∈ Vi0 . Since M is nilpotent, there exists a path p = (i1 |αhn | . . |αh1 |i0 ) in Q such that p · v = 0 but (α · p) · v = 0 for any arrow leaving the vertex i1 . But this means that S(i1 ) M is a non-zero subrepresentation. We infer that M = S(i1 ) by simplicity of M . 20. Let Q be a quiver. The dimension of the space Ext1Q (S(i), S(j)) equals the number cij of arrows in Q from i to j. 1 Generalities on quivers Proof. 14. When i = j, the domain of the map c ≡ cS(i),S(j) : Homk (S(i)s(α) , S(j)t(α) ) ∼ = Homk (S(i)l , S(j)l ) −→ l∈Q0 Homk (k, k) α:i→j α∈Q1 is the zero-dimensional vector space.

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