A First Course of Homological Algebra by D. G. Northcott

By D. G. Northcott

In keeping with a chain of lectures given at Sheffield in the course of 1971-72, this article is designed to introduce the scholar to homological algebra heading off the flowery equipment often linked to the topic. This ebook offers a couple of very important subject matters and develops the required instruments to address them on an advert hoc foundation. the ultimate bankruptcy comprises a few formerly unpublished fabric and may offer extra curiosity either for the prepared pupil and his educate. a few simply confirmed effects and demonstrations are left as workouts for the reader and extra workouts are incorporated to extend the most subject matters. options are supplied to all of those. a quick bibliography presents references to different guides within which the reader could stick to up the topics taken care of within the ebook. Graduate scholars will locate this a useful direction textual content as will these undergraduates who come to this topic of their ultimate yr.

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But gxv = 0 = g2v, and now (2) gives a contradiction. Exercise C. Let :A-+B and i/r:B^C be A-homomorphisms. ->Ci = 0; 22 THE LANGUAGE OF FUNCTORS (b) whenever there is a A-homomorphism d:X->B such that i/rd = 0, then 6 = ffor a unique A-homomorphism f\X-> A. Solution. Assume that (a) and (b) both hold. Then I m 0 c Ker^. KeTi/r^B is the inclusion mapping, then xjrj = 0 and therefore, by (6), j = $y for some A-homomorphism y:TLev\jr->A.

E. which ensures that HomA (n, B) (g) = / . Thus Ker{HomA(0-, B)} c; Im{HomA(77, B)} and the solution is complete. Exercise 2. Suppose that modules A, B belong to ^ A and that Y is the centre of A. Since A is a (A, Y)-bimodule, Hom^ (A,B) is a Y-module. Since B is a (A, Y)-bimodule this also induces a Y-module structure on HomA (A,B). Show that these two structures coincide. Solution. Let fe HomA (A, B) and y e Y. If g is the product of y and / i n the case where A is considered as a (A, F)-bimodule, then y 40 THE HOM FUNCTOR is a commutative diagram.

Let 2 consist of all the submodules of E that are essential extensions of A and let 2 be partially ordered by inclusion. Then A E 2 and, by Lemma 4, 2 is an inductive system. Zorn's Lemma now shows that 2 contains a maximal member C say. It will suffice to prove that C is injective and, by Theorem 16, this will follow if we show that G has no non-trivial essential extensions. Let D be an essential extension of G. Since E is injective, the inclusion mapping C^E can be extended to a A-homomorphism rf>:D->E.

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