# Tag Info

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Spanier's book is relatively old (so I know it does not quite answer your question), but excellent. It uses category theory from the get-go. Riehl's "Categorical homotopy theory" is very well-written, though it may be a bit too advanced if you hadn't seen a bit of algebraic topology already. Riehl's book is focused on the categorical aspect via Quillen model ...

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I suggest Peter May's Concise course on algebraic topology. You will find e.g. categorical formulations (and proofs) of the van Kampen theorem and the classification of covering spaces.

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The closest thing I've found is Strom's Modern Classical Homotopy Theory, although I haven't read much of it. Chapter 1 is called Categories and Functors, so that's a good start. This is the only introductory algebraic topology textbook I know of that explicitly uses the language of homotopy limits and colimits.

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You should embed the kernel $K$ into some injective module $Q$ and then apply $\operatorname{Hom}(-,Q)$ to the composition $$K \to A \to B.$$ You get a map which is both surjective and zero, hence $\operatorname{Hom}(K,Q)=0$. By our choice we have an embedding $K \subset Q$, so we deduce $K=0$.

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To prove that an $R$-module $M$ is flat, it suffices to show that for every ideal $I \subset R$, the canonical map $I \otimes_R M \rightarrow M$ is injective. When $R$ is a domain and $M$ is the field of fractions of $R$, we have that every element of $I \otimes_R M$ is expressible as a simple tensor, that is, $i \otimes m$ for some $i \in I$ and $m \in M$. ...

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A counterexample follows in the category of abelian groups. Let $A=\bigoplus_{n=0}^{\infty} \Bbb{Z}/2 \Bbb{Z}$, $B = \Bbb{Z}$. Define $g: B \oplus A \longrightarrow A$ to be the map $$(z, x_0, x_1,x_2, \dots) \mapsto (z \bmod{2}, x_0, x_1, x_2, \dots)$$ and $f: B \longrightarrow B \oplus A$ to be the map $$z \mapsto (2z, 0, 0, 0, \dots)$$ Then $0 \to B ... 2 The category$\mathcal{O}$is closed under quotienting, submodules and finite direct sums, but not under extensions. Hence$Y$need not be in$\mathcal{O}$in general (see here). 2 Take$R=\mathbb{Z}$,$C=0, C'$the standard extension of$\mathbb{Z}/2$by$\mathbb{Z}$, and$g$the identity of$\mathbb{Z}/2$in grade 0. Then$C\otimes D=0$but$C'\otimes D'$has homology$\mathbb{Z}/2$and$f\otimes g=0$. So$f\otimes g$need not be surjective on homology, and running the same example in the other direction shows it need not be ... 2 What is true is that$\rm Tor$commutes with filtered colimits. We already know that$M\otimes -$commutes with colimits, now take a right$R$-module$M$and a system$(N_i,\psi_{ji})$of left$R$-modules over some filtered set$I$. First, we can obtain a short exact sequence of filtered systems $$0\to (K_i,\rho_{ji})\to (P_i,\tilde\psi_{ji})\to ... 2 Consider the following commutative diagram of \Omega-modules with exact rows.$$\require{AMScd} \begin{CD} 0 @>>> I_k \otimes_k \Omega @>>> k[x_1,\ldots,x_n] \otimes_k \Omega @>>> (k[x_1,\ldots,x_n]/I_k) \otimes_k \Omega @>>> 0 \\ \ @VVV @VV\sim V @VVV \ \\ 0 @>>> I @>>> \Omega[x_1,\ldots,x_n] ... 2 Rotman's An Introduction To Algebraic Topology is a great book that treats the subject from a categorical point of view. Even just browsing the table of contents makes this clear: Chapter 0 begins with a brief review of categories and functors. Natural transformations appear in Chapter 9, followed by group and cogroup objects in Chapter 11. The aspect I ... 2 I don't think so. We could for example take$N=G={\mathbb Z}$and$T={\mathbb Z}/n{\mathbb Z}$for some$n>1$. Then$H^{2k}(T,{\mathbb Z}) ={\mathbb Z}/n{\mathbb Z}$, but all other terms in the sequence are$0$. 2 The homotopical significance of long exact sequences has to do with fiber sequences and their categorical duals, cofiber sequences. This is a long story and I don't know where it's told well and for students but the prototypical example for spaces is the following. If$f : E \to B$is a map of spaces$e \in E, b \in B$are basepoints such that$f(e) = b$, ... 2 One reason is this leads to modules that behave poorly with respect to commutators. For example, let$m\in M$,$n\in N$, and suppose$r,s\in R$don't commute. Then $$rm\otimes sn=rs(m \otimes n)=sr(m \otimes n)$$ So any commutator$rs-sr$annihilates the whole tensor product. 1 Neither. The$\mathbb{Z}$-module$\mathbb{Q}$is the direct limit of its finitely generated submodules. A finitely generated submodule of$\mathbb{Q}$is actually infinite cyclic and so free. However$\mathbb{Q}$is not projective. So the claim is false even for filtered direct limits. For inverse limits of injective modules, see this paper by Bergman on ... 1$\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Ext}{Ext}$I must have a different edition than you do, here's the statement of Theorem 3.4.3 that I read: Theorem 3.4.3. Given two$R$-modules$A$and$B$, the mapping$\Theta : \xi \mapsto \partial(\operatorname{id}_A)$establishes a 1-1 correspondence $$\{\text{equivalence classes of extensions of ... 1 This is a consequence of Tensor-Hom adjunction. In particular it suffices to assume that A is flat, which is slightly weaker than projective. 1 If F:R-\textrm{Mod}\to R-\textrm{Mod} is an exact additive covariant functor, then F commutes with (co)homology, that is, for every chain complex K_\bullet we have F(H_n(K_\bullet))\simeq H_n(F(K_\bullet)). Let$$K_\bullet: \cdots\to K_{n+1}\stackrel{f_{n+1}}\to K_n\stackrel{f_n}\to K_{n-1}\to\cdots0$$be a chain complex. Then ... 1 I'm basing myself on McCleary's book A user's guide to spectral sequences, sections 1.3 and 2.4. He actually calls them "spectral sequences of algebras", not "multiplicative spectral sequences". But it's probable that other sources give similar definitions (in the end, it all depends on what your applications are, I guess). Yes, the product is almost* ... 1 I think the following works. Let me know if anything is unclear; I've been trying for a while to also find an elementary proof of this fact, and I'll be very glad if this ends up being correct! We follow Eisenbud's hint. Suppose$\alpha\colon A \twoheadrightarrow B$and$\beta\colon B \rightarrowtail C$form such a factorization. Let$C\alpha$,$C\beta$be ... 1 In fact, a proof dual to the one you linked to works. Let's spell it out. Let$\ 0\to M\to I_0\to \ldots \to I_{n-1} \stackrel{f}{\to} I_n \to 0$be a minimal injective resolution of$M$. The module$I_n$is non-zero, since$M$has injective dimension$n$. Take$I=I_n$. To compute$Ext^n(I_n,M)$, we apply the functor$Hom(I_n,-)$to the above ... 1 Rotman's book: "An Introduction to Homological Algebra" has: And 1 I have no idea how Rotman gets from$\DeclareMathOperator{\pd}{pd}\pd_R(M)=n\lt\infty$that$\operatorname{Ext}_R^{n+1}(L,M)=0$for all$R$-modules$L$. (In my opinion this is a mistake.) Instead I can show you an alternative proof: Let$x\in Z(R)$be an element which is not a zero-divisor and let$R^*=R/(x)$. Moreover, let$M$be a left$R$-module with ... 1 It is a standard fact of homological algebra that derived functors can be computed from acyclic resolutions. In other words, if$F$is a left exact functor and$K^\bullet \to M$is a resolution by$F$-acyclic objects (ie.$R^iF(K^j) = 0$for all$i > 0$), then$R^iF(M) = H^i(F(K^\bullet))$. You can find a proof of this (for the left derived case, ... 1 Take an element$b\in\ker s^\to$, and an element$b_i$in some$B_i$such that$\beta_i(b_i)=b $. Then$\gamma_i(s_i(b_i))=0$, so that$\gamma_{ij}(s_i(b_i))=0$in some$C_j\enspace(j\ge i)$. As$\,\gamma_{ij}s_i=s_j\beta_{ij}$, this means$\,b_j=\beta_{ij}(b_i)\in\ker s_j$, so there exists$a_j\in A_j$such that$b_j=r_j(a_j)\$. Since ...

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If you have a short exact sequence of modules and localize it, it remains exact. What is the relation between localizing a module and tensoring it with the field of fractions?

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