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THE BASICS OF EXT AND TOR In lecture, we have omitted the proofs about some basic facts of Ext and Tor, including that they are well defined. Our reasoning is that the true interest of homological algebra is the application of these tools, and these proofs do not contribute much towards this aim. However, being mathematicians, such omissions are contrary to our nature, thus for completeness sake, these exercises outline the proofs of these basic properties. They will not be collected, so feel free to do these whenever you like (perhaps after the course is over). If you get stuck, their solutions can be found in any standard homological algebra book. We assume R to be a commutative ring, although this assumption is not strictly necessary. These results hold for general abelian categories with enough projectives, but are harder to prove. We do not assume finite generation or Noetherianism, but one may do so if they wish. (1) Given chain complexes, A• , B• , a chain map f : A• → B• is collection of R-module homomorphisms fn : An → Bn such that fn−1 ◦ ∂nA = ∂nB ◦ fn . a) Show that the set of chain maps C(A• , B• ) is an R module. Note that chain complexes along with chain maps form a category, denoted Ch(R) (what needs to be checked to show this?). b) Show that a chain map f induces homomorphisms f˜n : Hn (A• ) → Hn (B• ). Hence Hn is a functor from Ch(R) to Mod R (what needs to be checked to show this?). (2) Two maps f, g are homotopic, written f ∼ g, if there exists a set of homomorphisms σn : An → An+1 such that: B fn − gn = ∂n+1 ◦ σn + σn−1 ◦ ∂nA Note that we call σ a homotopy from f to g. a) Show that homotopy is an equivalence relations and that the set of chain maps modulo homotopy, i.e. C(A• , B• )/ ∼ is an R module. Note that chain complexes with along with equivalence classes of homotopic maps forms a category which we call the homotopy category and denote by K(R) (what needs to be checked to show this?). Furthermore, there is a natural functor from Ch(R) → K(R). b) Two chain complexes A• and B• are homotopic if there exists chain maps from f : A• → B• and g : B• → A• such that gf ' idA• and f g ' idB• where ' denotes homotopy equivalence. Show that homotopy of chain complexes is an equivalence relation on chain complexes. Note, that this follows from the previous note, since chain complexes are homotopic precisely if they are isomorphic in K(R). c) Show that homotopic chain maps induce the same map on homologies. Hence Hn is a functor from K(R) to Mod R (what needs to be checked to show this?). d) Conclude that homotopic chain complexes have the same homologies. (Is the converse true?) (3) a) Show that any module homomorphism f : M → N and any projective resolutions P• and Q• of M and N , f lifts to a chain map between P• and Q• respectively. (Hint: Use the definition of projective) b) Furthermore, show that such a lift is unique up to homotopy. c) Show that for any module M , all projective resolutions of M are homotopic as chain complexes. Hence projective resolutions are unique in the category K(R). Furthermore, taking a projective resolution is a functor from Mod(R) to K(R). (4) Let X be a module. a) For a complex A• , show that the modules X ⊗ An and differential X ⊗ ∂nA defines a chain complex, which we will denote by X ⊗ A• . b) Show for a chain map f : A• → B• that X ⊗ f = {X ⊗ fn } is a chain map from X ⊗ A• to X ⊗ B• . Furthermore, show that if f, g are homotopic, then X ⊗ f and X ⊗ g are homotopic. Note that this shows that X ⊗ − : K(R) → K(R) is a functor c) Show that Tori is well defined. Furthermore, show that it is functorial, i.e. given an f : M → N , there is a well defined homomorphism Tori (f, X) : Tori (M, X) → Tori (N, X). d) Rework the exercise using the functor Hom(−, X). 1 2 THE BASICS OF EXT AND TOR α β (5) A short exact sequence of chain complexes is a sequence of chain maps 0 → A• − → B• − → C• → 0 such α βn n Bn −−→ Cn → 0 is an exact sequence. Show that short exact sequence of chain that each 0 → An −−→ maps yields a long exact sequence of homologies: δn+1 α βn δ n n Hn (A• ) → · · · Hn (B• ) −−→ Hn (C• ) −→ · · · → Hn+1 (C• ) −−−→ Hn (A• ) −−→ (Hint: You will need to look up and apply the snake lemma.) (6) (Horseshoe Lemma) Given a short exact sequence of modules 0 → M → N → L → 0 and projective resolutions P• and P•0 of M and L respectively, show that there exists a projective resolution P•00 of N such that 0 → P• → P•00 → P•0 → 0 is exact. (Hint: Pn00 = Pn ⊕ Pn0 , although the differential will in 0 general not be ∂ P ⊕ ∂ P ) (7) Show that for a short exact sequence 0 → M → N → L → 0 and any module X, we get the long exact sequences: · · · Tor2 (N, X) → Tor2 (L, X) → Tor1 (M, X) → Tor1 (L, X) → Tor1 (N, X) → M ⊗ X → N ⊗ X → N ⊗ X → 0 0 → Hom(N, X) → Hom(L, X) → Hom(M, X) → Ext1 (N, X) → Ext1 (L, X) → Ext1 (M, X) → Ext2 (N, X) → · · · (8) For an element x ∈ R, define µ : M → M by m 7→ xm. Show that Tori (µ, X) and Exti (µ, X) are also given by multiplication by x.