# Download PDF by Tomi Pannila: An Introduction to Homological Algebra By Tomi Pannila

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Let k1 : ker f Ñ f and k2 : ker e Ñ e be the corresponding morphisms. Then f k1 “ 0 implies ek1 “ 0 and ek2 “ 0 implies f k2 “ 0. From the universal properties if follows that ker f – ker e. The rest of this section is devoted for a quick introduction to exact sequences in abelian categories. 13 (Exact sequence). Let A be an abelian category. A sequence of objects and morphisms of the form ... i´1 φi´2 X i´1 φi´1 i φi Xi X i`1 i´1 φi`1 ... i´1 is an exact sequence if Im φ – ker φ , where Im φ “ kerpcoker φ q, for all i P Z.

An epimorphism f : A Ñ B is strong if for every commutative square A g C f B h g1 f1 D 1 where f is a monomorphism, there exists a unique morphism h : B Ñ C making the diagram commutative. An epimorphism is said regular if it is the coequalizer of some pair of morphisms. 8. A regular epimorphism is a strong epimorphism. Proof. Let f be the coequalizer of a, b : A Ñ B such that the following diagram is commutative a,b A f B g f1 X C h g1 Y where f 1 is a monomorphism. By commutativity g 1 f a “ f 1 ga and g 1 f b “ f 1 gb.

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. We have f pa1 q “ 0 because ppf pa1 qq “˚ gpk2 pxqq “ 0 and p is a monomorphism. Thus, there exists a pseudo-element y P˚ ker f such that k1 pyq “˚ a1 . Since k2 is a monomorphism, f˜1 pyq “˚ x. 3 (iv). Let w P˚ ker g. 3 (ii) 0 P˚ A2 is the unique pseudo-element, up to pseudo-equality, with pseudo-image gk2 pwq. 13) we get ˜g1 pwq “˚ c1 p0q “ 0.