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9

0) The excellent mathematician you evoke has as family name (=surname) tom Dieck and as first name Tammo: tom is part of his surname and has nothing to do with Tom, the endearing form of Thomas. 1) Your idea of "finding a book that starts with a formal treatment of the basics of category theory and moves to more advanced/specialized concepts in a ...

7

A category with this property is called balanced, although I think this term is terrible. In practice, the problem that arises is that the monomorphisms in some category correspond to our intuitive notion of the "injective" maps, but the epimorphisms frequently fail to correspond to our intuitive notion of the "surjective" maps; for example, in ...

5

If $m = n$, then $0 \to \mathbb{Z}/n\mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z}/n\mathbb{Z} \to 0$ is a free resolution. If $m < n$, it will be useful to write $n = km$, where $k > 1$. The first step of the resolution looks like $F_0 \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$ for some free $\mathbb{Z}/n\mathbb{Z}$-module $F_0$. ...

4

Let $\mathbb{Z}_p$ denote the $p$-adic integers. Then $$Ext(\mathbb{Q},\mathbb{Z}_p)=0$$ satisfies the requirements in the question. Before giving the proof, let's put the example into a broader context: An abelian group $G$ is called cotorsion, if $Ext(F,G)=0$ for all torsion-free abelian groups $F$ (actually this is equivalent to ...

4

You have $\text{Hom}(A,G(-))=\text{Hom}(A,-)\circ G$, so you need exactness of both $\text{Hom}(A,-)$ (i.e., projectivity of $A$) and $G$ to ensure exactness of the composition.

4

$\require{AMScd}$ As noted above, the first argument provides an injection in homology, but not an isomorphism. I will set things up inductively so that we get a diagram $$\begin{CD} \\ &{}&{}& C_n @>>> C_{n-1} \\ {}&{}& {} @Af_n AA @A A f_{n-1} A \\ {}&{}&{}&P_{n} @>>d_n> P_{n-1} \\ \end{CD}$$ ...

3

I'm a bit puzzled by the use of the term "chain homotopy", as a chain homotopy is something you have between chain maps, not between complexes. It's certainly not true in general that the complexes $F\mathcal{A}_*$ and $F\mathcal{I}_*$ are chain homotopy equivalent, but what is true is that they have the same cohomology, and so either can be used to ...

3

If $k$ is a field, and your quiver has no oriented cycles, then there are exactly as many indecomposable projective modules (up to isomorphism) as there are vertices in the quiver. Moreover, any finite-dimensional representation admits a projective cover. (This is all true because of more general results on Artinian algebras). In your specific example, to ...

3

Not necessarily. For instance, the shift functor is an automorphism of any triangulated category, but it usually does not preserve distinguished triangles (because the shift of a distinguished triangle need not be distinguished, only the shift with the signs of all the maps reversed is distinguished). For a general discussion of equivalences of ...

3

This is not necessarily true. Consider the following bicomplex: $$K^{p,q} = \begin{cases} \mathbb{Z}, & q=-p \text{ or } q=-p+1; \\ 0, & \text{otherwise.} \end{cases}$$ The differentials are defined as follows: $d : K^{p,-p} \to K^{p,-p+1}$ (the vertical maps) is multiplication by $2$, while $\delta : K^{p,-p} \to K^{p+1,-p}$ (the horizontal maps) ...

3

You can't solve them because the statements are false. For (b) the counterexample is the Prüfer $p$-group $\mathbb{Z}(p^\infty)$, for which $$\mathbb{Z}(p^\infty)/A\cong \mathbb{Z}(p^\infty)$$ for every proper subgroup $A$. Of course this also gives counterexamples to (a), because $\mathbb{Z}(p^\infty)$ has infinitely many subgroups and they form a ...

3

Take $R=\mathbb{Z}$, $M=\mathbb{Q}\oplus \mathbb{Z}/2\mathbb{Z}$ and $N= \mathbb{Z}/2\mathbb{Z}$, with inclusion $R\to M:n\mapsto (n,\overline{0})$. Then, in $M\otimes_\mathbb{Z} N$, we have that $$(1,\overline{0})\otimes \overline{1} = (\frac{1}{2},\overline{0})\otimes 2\cdot\overline{1} = (\frac{1}{2},\overline{0})\otimes \overline{0} = 0.$$ Therefore ...

2

First, note that there always exists a left-$add(X)$-approximation $\varphi:C\to X^{\widehat{m}}$ for any module $C$ in mod-$A$. This comes from the fact that $Hom_A(C,X)$ is a finitely generated module over the (implicit) base ring $k$; if $f_1, \ldots, f_r$ are generators, then the map given in matrix form by $(f_1, \ldots, f_r)^t$ from $C$ to $X^r$ is an ...

2

The original sequence being exact at $X^n$ is equivalent to the map $\theta:C\to X^m$ induced by $\beta$ being injective. If $\varphi:C\to X'$ is any left $\text{add}(X)$-approximation, then $\theta$ factors through $\varphi$, and so $\varphi$ must also be injective, and the resulting sequence $0\to Y\to X^n\to X'$ is exact.

2

Concerning the third question the answer is yes. Suppose that two short exact sequences are equivalent. Then the middle groups are isomorphic by using the $5$-lemma (the morphism between the middle groups is injective and surjective). The converse need not be true. An example here are the $p$ inequivalent extensions of $C_p$ by $C_p$, while there are only ...

2

Here is a counterexample: Consider the category with three objects $P$, $X$ and $Y$, and two non-identity morphisms $f : P \to X$ and $\pi : Y \to X$ (and the only possible composition rules). Then every epimorphism with codomain $P$ splits (there is only one such epimorphism, the identity...). But $P$ is not projective: $\pi$ is an epimorphism, but $f$ does ...

2

I would suggest the following proof which one can find it in Rotman's book (An Introduction to homological algebra p134). I also assume $_{\mathbb{Z}}N$ is f.g as you used it like that in your proof. Let $N$ be a f.g $\mathbb{Z}$-module, since $N$ is torsion free hence it is free, by using the fundamental theorem of finitely generated modules over PID. ...

2

The problem of understanding elements in $\mathcal{K}_0(\mathcal{D}^b(\mathcal{A}))$ can be somewhat reduced to that of understanding elements in $\mathcal{K}_0(\mathcal{A})$ via the isomorphism $$\mathcal{K}_0(\mathcal{D}^b(\mathcal{A})) \to \mathcal{K}_0(\mathcal{A}): [X] \mapsto \sum_{i=-\infty}^{\infty}(-1)^i [H^i(X)].$$ Thus if you know the cohomology ...

2

Yes, that is true - besides than the fundamental observation that there's no reason why the specific elements of a group would matter, so any isomorphic copy of a group is as good as any other for this purpose, you can make an obvious bijection between SES's $$0\to A\to X\to B\to0$$ and SES's $$0\to C\to X\to D\to0$$ using the isomorphisms $A\cong C$ and ...

2

I) More generally, consider a short exact sequence $0\longrightarrow I\xrightarrow{\ \ \iota\ \ }R\xrightarrow{\ \ \pi\ \ }R/I\longrightarrow 0$. Tensoring with $M$ gives a right exact sequence $$I\otimes M\xrightarrow{\iota\otimes\operatorname{id}}R\otimes M\xrightarrow{\pi\otimes\operatorname{id}} (R/I)\otimes M\longrightarrow 0\,.$$ Now there is an ...

1

Let $I$ be a left ideal of $R$ and $M$ be a right $R$-module. You can consider the commutative diagram with exact rows $$\require{AMScd} \begin{CD} {} @. M\otimes_R I @>>> M\otimes_R R @>>> M\otimes_R R/I @>>> 0 \\ @. @VVV @VVV @VV\varphi V \\ 0 @>>> MI @>>> M @>>> M/MI @>>> 0 \end{CD}$$ where ...

1

"$\Leftarrow$" If $a\in A$ annihilates $M$, then $a$ annihilates $M^r$. In particular, $a$ annihilates $A$ and hence $a=0$. "$\Rightarrow$" If $(x_i)_{i\in I}$ is a generating set for $M$, then there is an injective homomorphism $A\to\prod_{i\in I} Ax_i$ given by $a\mapsto(ax_i)_{i\in I}$, and thus we have an exact sequence $$0\to A\stackrel{f}\to M^I.$$ If ...

1

Let $k$ be a field. It can be shown that in the category of sheaves of $k$ vector spaces, a constant sheaf $M_X$ is injective if and only if it is flabby (see Sheaves on Manifolds, Kashiwara-Shapira, exercise II.10, p. 133). Now, if $X$ is an irreducible noetherian space, for exemple any irreducible complex algebraic variety endowed with the Zarisky ...

1

Let's setup the framework. Let $\mathsf{Ab}$ be the category of abelian groups, and let $\mathsf{SES}$ be the category of short exact sequences, that is diagrams of the form $0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$ (with the usual conditions about images and kernels) and morphisms the obvious thing. Then define three functors, \begin{align} ... 1 Apparently, "minimal faithful" is meant to mean "minimal among faithful ideals." Let R be some interesting commutative quasi-Frobenius ring (say, \Bbb Z/(36) ). In such a ring, nontrivial ideals have nontrivial annihilators due to the double-annihilator property. This means that the entire ring is a minimal faithful ideal (actually, the minimum faithful ... 1 Homological Algebra is a game of arrows, if you can fully make use of the arrows, you can prove theorems. First of all, let us define \sigma_{AB}, \sigma_{AD}, \sigma_{BD}, \sigma_{BC}, \sigma_{CE}, \sigma_{DE} be maps of corresponding arrows. LetB_n \xrightarrow{\phi} C_n \oplus D_n\xrightarrow{\epsilon}E_n$$with$$\phi=\sigma_{BC} \oplus ...

1

You're indeed on the right track. $x_1$ already has the correct image in $C_n$, but unfortunately maybe not in $D_n$, so we need to correct $x_1$ with something, such that the image in $C_n$ stays untouched and the image in $D_n$ is the right one. Call the image of $x_1$ in $D_n$ $d$. The image of $d$ in $E_n$ is $x'=-y'$, hence $d+y$ is zero in $E_n$, so ...

1

You forgot the functor $\operatorname{Hom}_R(\text{--},M)$ is contravariant. Let $0\longrightarrow N'\xrightarrow{\ u\ }N$. What has to be proved is $$\mathrm{Hom}_R(N,\widetilde{M})\rightarrow \mathrm{Hom}_R(N',\widetilde{M})$$ is surjective. For this, note first \operatorname{Hom}_R(N,\widetilde{M})=\operatorname{Hom}_R(N,\operatorname{Hom}_\mathbf ...

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