By Tomi Pannila

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**Additional info for An Introduction to Homological Algebra**

**Sample text**

3 (vi) there exists a pseudo-element ˚ x0 P X1 such that k 1 px0 q “˚ y and e1 px0 q “˚ x. Now ˜ p1 δpxq “˚ φpxq “˚ u1 gk 1 px0 q “˚ u1 ppzq “˚ p1 upzq. 3 (ii) we have that δpxq “˚ upzq. Exactness at ker f : We need to show that f˜1 is a monomorphism. But this follows from the equality k2 f˜1 “ mk1 , because k2 , m, and k1 are monomorphisms. 3 (iv). By commutativity and the fact that k3 is a monomorphism, g˜1 f˜1 is the zero morphism. Let x P˚ ker g such that g˜1 pxq “ 0. Now, ek2 pxq “ 0, so by exactness of the second row, there exists a pseudo-element a1 P˚ A1 such that mpa1 q “˚ k2 pxq.

This implies that f a1 p is an epimorphism and hence f a1 is an epimorphism. Now xf a1 “ 0, so x “ 0. Thus f is an epimorphism. 10. Since gf “ 0, we have gf paq “ 0 for all pseudo-elements a of A. Let b P˚ B such that gb “ 0. Then there exists a unique morphism 34 c such that mc “ b because by exactness m “ ker g. 5. By commutativity f a “ qb, so f paq “˚ b. 3) is exact, let f “ me be the epimorphism monomorphism factorization of f . It suffices to show that m “ ker g. Now gf “ 0, so gm “ 0 because e is an epimorphism.

10 applied to the morphisms A1 Ñ B1 and B2 Ñ C2 . 10 we get unique morphisms A11 Ñ A2 and C1 Ñ B21 which keep the diagram commutative. Let X11 be the kernel of C1 Ñ B21 and X21 the cokernel of A11 Ñ B2 . It is not hard to see that we obtain factorizations β “ β2 β1 and τ “ τ2 τ1 through the kernel X11 and the cokernel X21 . Let us denote by X1 the pullback of the morphisms η and e and by X2 the pushout of the morphisms p and A2 Ñ X21 . 2 and are the kernel and the cokernel of X1 Ñ X11 and X21 Ñ X2 , respectively.