# Tag Info

5

For question 2, $F\otimes_A -$ has a left adjoint iff $F$ is finitely generated, and the left adjoint is always exact. For if $(f_i)\in F^I$ is an element of an infinite product of copies of $F$, then it is easy to see $(f_i)$ is in the image of the canonical map $F\otimes_A A^I\to F^I$ iff $\{f_i\}$ is contained in a finitely generated submodule of $F$. ...

5

You can take homology $H_{\bullet}(X, A)$ with coefficients in any abelian group $A$. Integral homology is the special case where $A = \mathbb{Z}$. "Homology" with no qualifier usually refers to this.

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Let $r,p:Z\to W$ be morphisms such that $r\circ g=p\circ g$. Then $r\circ g\circ f=p\circ g\circ f$, which implies that $r=p$ since $g\circ f$ is epi. Thus, $g$ is also epi. Note that I didn't use the abelianess of the category, so it holds in arbitrary categories.

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In general, all limits and colimits of $k$-algebras exist. Limits are taken set-theoretically. To describe colimits it suffices to describe coproducts and coequalizers. The coproduct of $k$-algebras is the free product, and the coequalizer of two maps $f, g : A \to B$ is given by quotienting $B$ by the two-sided ideal generated by elements of the form $f(a) ... 4 Intuitively, the isomorphism preserves products because just as the cup product comes from pulling back Künneth via the diagonal, the product seems to come from pulling back, i.e. dualizing via$\text{Hom}$, a derived version of the "Künneth"$\mathbb{Z} \cong \mathbb{Z} \otimes \mathbb{Z}$. 4 These categories are not just equivalent but isomorphic. You have defined an operation on objects$Mod(A)\to Mod(\tilde{A})$, and it is easy to see that it also preserves maps and so gives a functor; the inverse is simply given by taking an$\tilde{A}$-module and restricting its module structure to$A\subset\tilde{A}$. More generally, this works with ... 4 We can compute$H_i(S^n\times X)$for any space$X$and all$i$without using Künneth's theorem as follows: Step 1: There is a retraction$r:S^n\times X\to \{x_0\}\times X=X$given by pinching$S^n$. This means that$r\circ i=id_X$where$i:X=\{x_0\}\times X\hookrightarrow S^n\times X$is the inclusion. Then, by functoriality of$H_i$, the inclusion is ... 3 A common source of half-exact functors which are neither right-exact nor left-exact is derived functors, which are always half exact by the long exact sequences relating them but are rarely right-exact or left-exact. For instance, if$n>0$and$A$is any object of projective dimension$>n$, then the functor$\operatorname{Ext}^n(A,-)$is half-exact ... 3 If$Q$is a projective module and$f : Q \to S$is a homomorphism, then there exists a map$f' : Q \to P_S$with$f = \pi \circ f'$with$\pi : P_S \to S$the canonical epimorphism. The important thing to note is that if$f$is surjective, then$\text{im}(f') + \ker(\pi) = P_S$. As$\ker(\pi)$is superfluous (it is the radical of$P_S$) we conclude that$f'$... 3 Let me remove most of your assumptions and work with an arbitrary module$M$over an arbitrary commutative ring$R$. We'd like to know when the functor$M \otimes_R (-)$has a left adjoint. The answer is iff$M$is finitely presented projective, in which case the left adjoint is$M^{\ast} \otimes_R (-)$where$M^{\ast} = \text{Hom}_R(M, R)$. You can ... 3 As Justin Young commented, there are many different ways of proving this (and which one is most comprehesible to you will depend on your background); here's one. Write$T_n(A)=H_n(K(\pi,1),A)$; then$T_n$is a functor from the category of$\mathbb{Z}[\pi]$-modules to the category of abelian groups. Moreover, these functors together canonically have the ... 3 Yes, they are naturally isomorphic. Define$f:B_*\otimes C_*\to C_*\otimes B_*$by$f(b\otimes c)=(-1)^{|b||c|}c\otimes b$when$b$and$c$are homogeneous. We can compute $$f(d(b\otimes c))=(-1)^{|db||c|}c\otimes db+(-1)^{|b|}(-1)^{|b||dc|}dc\otimes b=(-1)^{(|b|-1)|c|}c\otimes db+(-1)^{|b||c|}dc\otimes b$$ and $$d(f(b\otimes c))=(-1)^{|b||c|}dc\otimes ... 3 I'm not sure what tools you have at your disposal, but you could try this: The short exact sequence 0 \to \mathbb{Z}\to \mathbb{Q}\to \mathbb{Q}/\mathbb{Z} \to 0 induces a long exact sequence of Tor-modules. In this long exact sequence is a piece which looks like$$\mathrm{Tor}_1^\mathbb{Z}(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \to ... 3 Let$f \in \operatorname{Hom}(\Bbb{Z}_p , \Bbb{Z})$. Then by the first isomorphism theorem,$\Bbb{Z}_p / \ker f$is isomorphic to a subgroup of$\Bbb{Z}$. Recall that nontrivial subgroups of$\Bbb{Z}$are infinite cyclic. But$\Bbb{Z}_p$has no infinite quotients, so necessarily$\Bbb{Z}_p / \ker f$is trivial! This implies that$\ker f = \Bbb{Z}_p$, i. e. ... 2 Here$[n] \overset{df}{=} \{1,\dots,n\}$means that$[n]$is defined to be$\{1,\dots,n\}$. Other common notations for this are$:=$,$\overset{def}{=}$(in fact I wouldn't be surprised if there was actually an 'e' missing),$\overset{\Delta}{=}$... 2 Yes of course this is true, and in fact, this is said in the wikipedia article you linked ! Indeed, in the exact sequence involving the$A_i$'s, the$C_i$do have a concrete description :$C_i=\ker(A_i\rightarrow A_{i+1})=\operatorname{im}(A_{i-1}\rightarrow A_i)$. So if you just consider$f:A_1\rightarrow A_2$, you get a short exact sequence $$... 2 The third property (with G=\mathbb{Z}) tells you that if C is a finite cyclic group, then \operatorname{Ext}(C,\mathbb{Z})\cong C. Furthermore, any finitely generated torsion abelian group is a direct sum of cyclic groups. So T is a direct sum of cyclic groups, so by the first and third properties, \operatorname{Ext}(T,\mathbb{Z})\cong T. Note, ... 2 To elaborate on my comment, applying \text{Ext}^{\bullet}(A, -) to the short exact sequence 0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0 produces the long exact sequence$$0 \to \text{Hom}(A, \mathbb{Z}) \to \text{Hom}(A, \mathbb{R}) \to \text{Hom}(A, S^1) \to \text{Ext}^1(A, \mathbb{Z}) \to 0$$where we can ignore the rest of the sequence because ... 2 Homology involves chains, boundaries, and cycles, right? And a typical cycle looks like$$ c_1 s_1 + c_2 s_2 + \ldots + c_n s_n $$where the coefficients c_i are in some group, and the s_i are simplices of some sort. If the group used is the integers, you get integral homology; if it's something else (e.g., \mathbb R) you get a different homology. ... 2 No. For example, \mathbb{Z}/p^2\mathbb{Z} has a composition series consisting of two copies of \mathbb{Z}/p\mathbb{Z}, but it is not isomorphic to (\mathbb{Z}/p\mathbb{Z})^2. As in your previous question, this reflects the existence of interesting Ext groups in general (even between simple objects). 2 As you said, free resolutions are projective resolutions. But it is important to remmember that in homological algebra you just need to take ANY projective resolution, in most cases. The big difference is that it is easier (in a way) to construct a free resolution than a projective resolution. Just look at the prove for the existense of enough projective ... 2 We have$$H_0=\frac{\ker\partial_0}{\operatorname{im}\partial_1}.$$Here, \ker\partial_0 is the free R-module generated by 1,2,3,4 and \operatorname{im}\partial_1 is the submodule of \ker\partial_0 generated by 2-1,3-1 and 4-1. Therefore the elements of the quotient have the form$$\alpha ... 2 This is just the commutativity of the restriction maps with the cup product on the level of cochains combined with the fact that$\tilde{H}_\bullet(X) = H_\bullet(X, *)$. 2 You really need to take a CW structure on$\bar X$which is inherited by a CW structure on$X$. This means a priori your CW structure is fixed during the whole construction, and in the end you show the independence of the resulting homology of the chosen structure (as you do in cellular homology as well). Fix a CW structure on$X$. The covering space ... 2 Suppose that$\mathbb{A}$is an abelian category and that$0\to K \to A\to B\to 0$is an exact sequence which splits. Let us write$\kappa: K\to A$,$\alpha:A\to B$and$\beta : B\to A$for the morphisms involved. By exactness we have that$\kappa$is the kernel of$\alpha$,$\alpha$is the cokernel of$\kappa$, and since the sequence splits$\alpha\beta ...

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The only ring homomorphisms $f:R\to S$ such that $f^*$ is an equivalence of derived categories are isomorphisms. So the answer to your question is that yes, $D(A)$ does have something to do with the categories $D(A_n)$: they're all equivalent, but for a trivial reason. [To prove my claim, suppose $f:R\to S$ is a ring homomorphism such that $f^*$ is an ...

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The terms that appear in the universal coefficient theorem are not just abelian groups; they are functors taking values in abelian groups, and the morphisms in the universal coefficient theorem are natural transformations. To ask for a natural splitting is to ask for a splitting which also organizes into a natural transformation.

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From the commutative diagram here, $H_\bullet(X, A \cup B) = H_\bullet(X, X) = 0$, so if we can show the left arrow is surjective, we are done. However, on each factor, it is just a restriction map, so this is clearly the case.

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EDIT: Jz Pan is correct; there is an issue with the fact that in the Proposition below, the map $Z_1 \to Z_2$ in the $3 \times 3$ diagram is no longer (necessarily) the same as the one we started with. We have the following: Proposition ([BBD, Prop. 1.1.11] or [Stacks, Tag 05R0]). Let $\mathscr{D}$ be a triangulated category. Given a commutative diagram ...

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More generally, if $F:Ab\to Ab$ is any functor which preserves addition of maps, then it sends split exact sequences to split exact sequences. This follows from the following theorem: Theorem Let $A\stackrel{i}{\to} B\stackrel{p}{\to} C$ be a pair of maps of abelian groups. Then the following are equivalent: There exist maps $q:B\to A$ and ...

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