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4

There seems to be a misunderstanding here. Your $\mathfrak M_Λ(F-, A)$ is not a functor from sets to "module homomorphisms", it is a contravariant functor from sets to sets, which assigns to every set $S$ a set of certain module homomorphisms. In other words, it is a functor from $\mathfrak S^\mathrm{op}$ to $\mathfrak S$, as is $\mathfrak S(-, GA)$. That ...


4

No, at least not in any obvious sense; it's not hard to give counterexamples showing that it does not preserve most limits, e.g. equalizers or pullbacks, or for that matter infinite products. (Depending on where you take these limits you run into further problems, e.g. the homotopy category of topological spaces itself has very few limits.) Abstractly, ...


3

The standard construction is via a push-out. Consider an exact sequence $0\to K\to P\to C\to 0$, with $P$ projective. Applying $\operatorname{Hom}_R(-,A)$, we get the exact sequence $$\def\H{\operatorname{Hom}_R}\def\E{\operatorname{Ext}_R^1} 0\to\H(C,A)\to\H(P,A)\to\H(K,A)\xrightarrow{\delta}\E(C,A)\to0 $$ Take $f\colon K\to A$ such that $\delta(f)=x$, so ...


3

Take $C_*=0\to \mathbb{Z}\stackrel{2}\to\mathbb{Z}\to 0$ (with the $\mathbb{Z}$s in degree $0$ and $1$ and every other term in the complex $0$) and let $D_*=0\to\mathbb{Z}/2\mathbb{Z}\to 0$ (with the $\mathbb{Z}/2\mathbb{Z}$ in degree $0$ and every other term in the complex $0$). Then there is a unique nonzero chain map $f:C_*\to D_*$, and this induces an ...


3

Yes, this is true over $\mathbb{Z}$ by the universal coefficient theorem over $\mathbb{Z}$. For the question with "homology" and "cohomology" switched see this math.SE question.


3

It translates as (at least mathematically. I have changed from the notation $x^y$ to $y(x)$ and changed to some more common letters for the various things): a) Let $N\unlhd G$ and $\alpha$ be an automorphism of $G$ such that $\alpha(n) = n$ for all $n\in N$ and $\alpha(g)N = gN$ for all $g\in G$. Then $\alpha(g) = gF(gN)$ for $1$-cocycle $F$ in ...


3

These conditions are equivalent if $X$ is a levelwise finite CW complex (finitely many cells of each dimension), since this condition ensures that each homology group is finitely generated. You can prove this using the universal coefficient theorem as described here; it ensures that the torsion subgroup of $H_k(X, \mathbb{Z})$ is isomorphic to the torsion ...


2

This is true if you assume $H_n(X)$ is finitely generated for all $n$ (all coefficients in this post will be $\mathbb{Z}$). In particular, this holds if $X$ has the homotopy type of a CW-complex with finitely many cells in each degree. To prove this, we invoke the classification of finitely generated abelian groups, which says that $H_n(X)$ is a finite ...


2

The extension is characterized by the triple $(i,B,p)$ (not by the isomorphism class of $B$ only) which is different of $(i,B,-p)$. The equivalence of extension is defined by this commutative diagram:https://en.wikipedia.org/wiki/Ext_functor#Equivalence_of_extensions not only by the isomorphism between $B$ and $B'$.


2

It's just a notation. It means "you can plug in an integer here and get something". It's also often used for simplicial sets, for example: a simplicial set $X_\bullet$ is a sequence of sets $(X_0, X_1, X_2, \dots)$ and some maps between them satisfying some identities. Or for example a cosimplicial set $Y^\bullet$ is also a sequence of spaces $(Y^0, Y^1, ...


2

This happens already for $\mathfrak{sl}_2$: The only non-semisimple block (up to equivalence) of BGG category $\mathfrak{sl}_2$ has two simple modules, $L$ and $S$ with $\operatorname{dim} \operatorname{Ext}^1(L,S)=\operatorname{dim}\operatorname{Ext}^1(S,L)=\operatorname{dim}\operatorname{Ext}^2(S,S)=1$. In this case, the category has global dimension $2$, ...


2

Let $\mathcal{A}$ be an abelian category and let $\mathcal{C}$ be the category of unbounded chain complexes in $\mathcal{A}$. For each integer $n$, we have a functor $(-)_n : \mathcal{C} \to \mathcal{A}$ sending a chain complex $C_\bullet$ to the object $C_n$, and it has both a right adjoint, namely the functor sending an object $A$ to the chain complex ...


1

Your guess is correct. I think that the easiest way to prove this is to note that for some $c,d$, $E^\bullet$ is quasi-isomorphic to a bounded complex of finitely generated projectives concentrated in degrees $[c,d]$ such that the first differential is not a split monomorphism and the last differential is not a (split) epimorphism, and that then $-d,-c$ are ...


1

A special case of completion is when $R$ is a commutative ring, $I$ is an ideal in $R$, and we complete $R$ with respect to $I$; this is the $I$-adic completion of $R$. It is a theorem (a consequence of Artin--Rees) that if $R$ is Noetherian, then any $I$-adic completion of $R$ is flat over $R$. In this case you can use results on Tor and flat base-change ...


1

The definition $P_i = \mathbb{Z}\left[G^{i+1}\right]$ should be made for all $i \geq -1$, not only for $i \geq 0$. Thus, $P_{-1} = \mathbb{Z}\left[G^{0}\right]$. But $G^0$ is the trivial group consisting of the one element $\left(\right)$ (the empty list, a.k.a. the $0$-tuple). If you plug in $i = 0$ into the definition of $d_i$, you obtain (1) ...


1

What you're missing is that the equation $d\alpha=0$ is taking place in $P_0/IP_0$, not in $P_0$. Backing up a bit, if $\alpha\in P_1/IP_1$ is a cycle, let's choose a lift $\beta\in P_1$ for $\alpha$. Then you can't necessarily say that $d\beta=0$ as an element of $P_0$, since all you know is that the image of $d\beta$ in the quotient $P_0/IP_0$ is $0$, ...


1

When computing $Ext^1(A,C)$, you are not only interested in how many $B$ can be fit into an exact sequence $$0\to C\to B\to A\to 0$$ but also, for a fixed $B$, how many distinct ways can you construct such a sequence. Of course, distinct means "distinct up to equivalence" as linked to in your previous question. That is, there is an isomorphism $\phi$ such ...


1

First, you mixed up exponents and indices quite liberally in this question... There's a difference. The map $f$ is a cocycle in $\operatorname{Hom}(X,Y)$ of degree $p+q$. So first, by definition of the degree in the chain complex $\operatorname{Hom}(X,Y)$, it will map a $p$-cochain of $X$ to a $q$-cochain of $Y$.* So since $z_2 \in Z^p(X)$, $f(z_2) \in ...


1

How does the uct arise? Because we have the natural epimorphism $H^k(M;G) \twoheadrightarrow Hom(H_k(M),G)$. UCT tells you that this splits, hence for $G=\mathbb Z$, as the kernel is torsion, you obtain for $k=3$ that this epimorphism is non-trivial. But as $H_3(M;\mathbb Z)$ is a finitely generated abelian group, $0\neq Hom(H_3(M),\mathbb Z) \to Hom(H_3, ...


1

The category of rings (or more generally the category of algebras over a ring, a ring being a $\mathbb{Z}$-algebra) has enough projectives. Let $A$ be a ring. Since the category of abelian group is known to have enough projectives, there is a projective abelian group $P$ and a surjective group morphism $q : P \to A$. Now consider the tensor algebra on $P$: ...


1

The following counterexample is available over a field $k$ other than $\mathbb{C}$. Let $A = B = M = N = L$ be a field extension of $k$. Then you want to ask (after the correction in the comments) whether $L \otimes_k L$ is a simple module over itself. For a commutative ring this is equivalent to being a field, but $L \otimes_k L$ is almost never a field. ...


1

It's probably far easier (and instructive) to prove the fact directly. (Moreover it's possible for a chain complex to have vanishing homology but not be contractible, consider the chain complex $$\dots \to \mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z} \xrightarrow{0} \dots;$$ but ...



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