# Tag Info

5

You don't show that $\mathcal{L}^{op}$ actually is a category of modules, just that any small subcategory of it fully and exactly embeds in one. Indeed, an abelian category with a projective generator is not usually equivalent to a category of modules: it only is if the projective generator is compact (and the category is cocomplete). In the case of $\... 4 As far as I know, there is no other translation. However, you should know that there were two translators, the other being Marcia L. Barr. She is a professional French/English translator and I put it into mathematicese. Very much against G's wishes; otherwise it would have appeared as TAC reprint. As for your question, in 1957 category theory wasn't even ... 4 In particular, for every$a \in A$it seems reasonable that$\mu_a : A \to A, \mu_a(b) = ab$should be a right DG-module homomophism. This is incorrect. The correct claim is that if$a$is a$0$-cycle then$\mu_a$is a dg module homomorphism. Explicitly, we have $$d(ab) = d(a) b + (-1)^{|a|} a \, d(b)$$ and so if$|a| = da = 0$then$d(ab) = a \, d(b)$, ... 4 The short exact sequence$0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \to \mathbb{Z}_2 \to 0$gives rise to a Bockstein exact sequence $$\cdots \to H_{i+1}(X, \mathbb{Z}_2) \to H_i(X, \mathbb{Z}) \xrightarrow{2} H_i(X, \mathbb{Z}) \to H_i(X, \mathbb{Z}_2) \to H_{i-1}(X, \mathbb{Z}) \to \cdots$$ from which it follows by exactness that the kernel of the map$...

3

The reason your search isn't successful is that when people talk about projective resolutions or free resolutions, there's not usually an implication that they are dealing with modules over a commutative ring. Standard homological textbooks develop homological algebra in more general categories, including the category of modules over a noncommutative ring. ...

3

The problem is that the "obvious" inclusion mapping $I \otimes B \to R \otimes B$, given by tensoring the inclusion mapping with the identity, need not be injective. Recall that, by definition, a module is flat if tensoring with that module preserves the head of exact sequences; but that is the same as saying that tensoring an injective homomorphism with ...

3

Yes, both homological dimensions agree. First, by Baer's criterion for injectivity, for any $A$-module $M$ you have $$\text{injdim}_A(M)=\max\{n\ |\ \text{Ext}^n_A(A/I,M)\neq 0\text{ for some }I\},$$ so you may restrict to finitely generated modules in the first argument when defining the homological dimension. For finitely generated modules $M$ (over a ...

3

Finite dimensional algebras have integrals, and using one you can show selfinjectivity at once —in fact, they are Frobenius. See for example the Lectures on Hopf algebras by Schneider, which Google will find for you, or Susan Montgomery's book in Hopf algebras.

2

I think it may be true if the ring is noetherian, but not for general rings. I'll build a counterexample in a few steps. First, there are easy examples of non-split short exact sequences $0\to A'\to A\to A''\to0$ of bounded complexes of finitely generated vector spaces over a field $k$. For example, there's an obvious such sequence with $$A'=\dots\to0\to0\... 2 For reasonable rings, one way to do this would be to use the fact that a module is projective if and only if it is locally so and thus we may assume A is local with residue field k. Now, M is projective if and only if \operatorname{Tor}^1_A(k,M)=0. Then one has 0=\operatorname{Tor}^1_B(k\otimes_A B, M\otimes_A B) =\operatorname{Tor}^1_A(k,M)\... 2 An obvious comment that nobody has mentioned: If you pass to a higher universe, then \mathrm{Mod}(R)^\mathrm{op} is small, and so the embedding theorem tells us that it embeds exactly into some \mathrm{Mod}(S). The catch is that (with respect to the original universe) S will be a large (i.e. proper-class-sized) ring. You don't actually need the ... 2 The trouble is that an object of the functor category [\cal B,\cal C] need not be projective even if every object in the image is. I think what you would get would be a weak homotopy (homotopy at each object of \cal B). For more on this subject, we my book titled Acyclic Models, which deals with every version of the theorem I am aware of. 2 If H_1(X;\mathbb Z) is finitely generated, you can calculate pretty explicitly what the map H_i(X;\mathbb Z) \to H_i(X;\mathbb Z) \otimes \mathbb Z_2 does--it takes everything to its value modulo 2. This is because H_1(X;\mathbb Z) \simeq \mathbb Z^k \oplus \bigoplus \mathbb Z_{p^i}^{d_{p^i}} by the structure theorem for finitely generated abelian ... 2 One source (which I have not read the whole article) might be the survey article "History of Homological Algebra" written by C. Weibel, http://www.math.uiuc.edu/K-theory/0245/survey.pdf. 1 After these references it really depends on what type of commutative or homological algebra you intend to work in. However, some of the most widely useful general references are as follows. Almost everything in the book Bruns and Herzog is lingua franca and can't be skipped. Most of the chapters of Weibel's Homological algebra. Here its I guess okay to ... 1 By the universal property of A \otimes_R B, there is a natural homomorphism A \otimes_R B \to (A \otimes B)/H, a \otimes_R b \mapsto a \otimes b + H. To go the other way around, note that the universal property of A \otimes B gives us a homomorphism A \otimes_R B \leftarrow A \otimes B; this map descends to a homomorphism A \otimes_R B \leftarrow (... 1 One example is the case R = \Bbb{Z}_4, B = I = \{0, 2\} that Osbourne is talking about at that point in the book. In this example, you find that I \otimes B \simeq R \otimes B \simeq \Bbb{Z}_2, but the natural mapping from I\otimes B to R \times B is not an injection. In fact, 2 \otimes 2 generates I \otimes B but maps to zero in R \otimes B, ... 1 Any map i : A \to B, not necessarily a monomorphism, induces a precomposition map \text{Hom}(B, C) \to \text{Hom}(A, C), and you want to take the cokernel of this map. If i is a monomorphism, it fits into a short exact sequence$$0 \to A \to B \to B/A \to 0$$which induces an Ext long exact sequence$$0 \to \text{Hom}(B/A, C) \to \text{Hom}(B, C) \...

1

Let me first give some names to your morphisms: $$\require{AMScd} \begin{CD} \cdots@>>>h^i(L^{\bullet}\otimes R)@>>>h^i(L^{\bullet}\otimes E)@>>>h^i(L^{\bullet}\otimes M)@>>>h^{i+1}(L^{\bullet}\otimes R)@>>>h^{i+1}(L^{\bullet}\otimes E)@>>>\cdots\\ @.@VV{\varphi^i_R}V@V\cong V{\varphi^i_E}V@VV{\varphi^i_M}... 1 Functoriality in the first variable only says that for k = 0 (morphisms have degree 0). If f is a morphism of degree k then it induces a map$$[X[k], Y] \to [W, Y] and there are some signs involved in moving this shift $[k]$ around.

1

Yes, the candidate is $BG$ the classifying space of $G$. You can endow $G$ with the discrete topology, you have the universal bundle $EG\rightarrow BG$ which is a covering and $EG$ is contractible $BG$ is a $K(G,1)$, apply Serre 1.5 p.91 Serre, Jean Pierre. Cohomologie des groupes discrets. http://www.maths.ed.ac.uk/~aar/papers/serrecoh2.pdf

1

If $0\to A^\bullet\stackrel{\alpha}{\to}B^\bullet\stackrel{\beta}{\to}C^\bullet\to0$ is a sequence of complexes over $\mathcal{A}$ such that, in every degree $m$, the sequence $0\to A^m\to B^m\to C^m\to0$ is a split short exact sequence, then there is a distinguished triangle $A^\bullet\stackrel{\alpha}{\to}B^\bullet\stackrel{\beta}{\to}C^\bullet\to A[1]$ in ...

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