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## Hot answers tagged homological-algebra

11

This boils down to: Must $n$ be the identity in this diagram? $$\require{AMScd} \begin{CD} A @>f>> B @>g>> C @>h>> D @>j>> E \\ @| @| @VVnV @| @| \\ A @>f>> B @>g>> C @>h>> D @>j>> E \\ \end{CD}$$ At least $c-n(c)\in\ker h$ for all $c\in C$, hence $c-n(c)=g(b)$ for some $b\in B$. Since ...

4

In fact all submodules of free $\mathbb{Z}$-modules are projective, even free, so the experimentation you're doing won't succeed. (This is because $\mathbb{Z}$ is a principal ideal domain.) Taking $(x,y)$ a submodule of $k[x,y]$ for $k$ some field will work better. $(x,y)$ is not projective, which we can show by showing it's not flat, since projective ...

2

Sometimes, you actually have to look at the topology. A chain $c$ in $H_1(X, A)$ is a collection of edges (with coefficients, but those will turn out to be irrelevant); the boundary is a collection of pairs-of-points in $A$. Since $A$ is path connected, for each such pair we can find a path in $A$ that connects them. The sum of all these paths, with the ...

2

As Olivier Begassat says in the comments, this will typically not be true because the fundamental group will typically fail to be abelian. For example, when $X = \mathbb{R}^2$, the fundamental group is the braid group $B_n$, which is nonabelian for $n \ge 3$. (In any case there isn't an obvious candidate for an H-space structure.)

2

No. Pick a field ${\mathbb k}$ and a finite oriented tree $Q$, considered as a category. Then the simple objects in the abelian functor category ${\mathscr A} := k\text{-Vect}^Q$ (which is called the category of representation of $Q$ and is equivalent to the category of modules over the path algebra ${\mathbb k}Q$ of $Q$ over ${\mathbb k}$) are in bijection ...

2

Differential Forms in Algebraic Topology by Bott and Tu has many examples that you could present if you want. A User's Guide to Spectral Sequences by McCleary has also tons and tons of examples. Here are some examples off the top of my head: The Serre SS for computing the homology of $\Omega S^n$ ("easy"), of $BU_n$ and $U_n$ ("hard"). In homological ...

2

Let $a' = \frac{a}{d}, n' = \frac{n}{d}.$ Then gcd$(a', n') = 1 \Rightarrow$ there exist $x, y \in \mathbb{Z}$ such that $a'x + n'y = 1.$ Note that $\{r = (-a', n'), s = (y, x) \}$ forms a $\mathbb{Z}$-basis of $\mathbb{Z} \oplus \mathbb{Z}$ (why?). Define a $\mathbb{Z}$-module homomorphism $\phi: \mathbb{Z}r \oplus \mathbb{Z}s \rightarrow ... 2 The definition in Cartan-Eilenberg is wrong. For,$t_i = T(A_1,\ldots, s_i,\ldots,A_r)$defines a map$T(A_1, \ldots, A_i,\ldots A_r) \to T(A_1, \ldots, A'_i,\ldots A_r)$. But for a homotopy you need a map$T(A_1, \ldots, A_r) \to T(A'_1, \ldots, A'_r)$. The correct definition of the homotopy in your case is$u=:(s \otimes g_1, f_2 \otimes t): A \otimes ...

2

The quasi-isomorphisms in the category of chain complexes are related to homotopy theory, and in a sense localizing at the quasi-isomorphisms is like localizing topological spaces at the homotopy equivalences. It is primarily a computational tool for chain complexes and resolutions of various objects. The purpose achieved by localizing, in general, is that ...

1

Every abelian category has fiber products. In fact, fiber products can always be constructed via equalizers and products, and equalizers in $\mathsf{Ab}$-categories may be reduced to kernels. Your question is why epimorphisms in abelian categories are stable under pullback. You can find the proof in Mac Lane's CWM, Chapter VIII, Prop. 4.2. By the way, you ...

1

By Theorem 9.1 (Chap. VI) in Hilton-Stammbach's A Course in Homological Algebra, $f$ induces for each $n$ an isomrphism $$f_n: G/G_n \to H/H_n.$$ Now let $x\in G$ represent an element from the kernel of $G/\bigcap_nG_n \to H/\bigcap_n H_n$. Hence $f(x)\in \bigcap_nH_n\subseteq H_n$. In particular $f_n(\bar{x})=\bar{1}$ and, by the theorem, ...

1

At the point where Lemma 1.2 is used you are already given maps $g_k: B_k\to M_k$ for $k<n$ such that $$(\alpha): \partial^M_k g_k = g_{k-1} \partial^B_k,\quad (\beta): j_k g_k = a_k\quad\text{and}\quad (\gamma): f_k g_k = b_k$$ hold, where $a_k: A_k\to M_k$ and $b_k: B_k\to N_k$ are the components of the given morphisms $A\to M$ and $B\to N$, ...

1

A periodic free resolution is a free resolution (that is, a resolution by free modules) which is periodic: if the resolution is $$\cdots\to P_n\xrightarrow{d_n}P_{n-1}\to\cdots\to P_0,$$ this means there exists a $p\geq1$ such that $P_{n+p}=P_n$ and $d_{n+p}=d_n$ for all $n\geq0$.

1

If you have two sections $s$, $s':G\to E$, then for all $g\in G$ we have that $s(g)$ and $s'(g)$ are two elements of $E$ with the same image in $G$: it follows that there exists a unique $t(g)\in A$ such that $s(g)=t(g)s'(g)$. Figure out what consequences this has in terms of the factor sets corresponding to $s$ and $s'$.

1

I think the 2nd quest is still open. By $s(x)s'(x)^{-1} \in \ker(p)=\text{im}(i)$ and injectivity of $i$ there is a unique $g(x) \in A$ such that $i(g(x))=s(x)s'(x)^{-1}$. This defines a function $g:G \to A$. Before we can deduce the formula in part (ii) of the question, it's important to note that the action of $G$ on $A$ is defined by $i(g\cdot a) := ... 1 For any ring$R$and any$n\geq 1$,$R$is Morita equivalent to$\text{Mat}_n\ R$through the$R-\text{Mat}_n\ R$-bimodule$R^{\oplus n}$. In particular, the relation$\sim$on central simple algebras induced by (i)$A\sim B$for$A\cong B$, and (ii)$A\sim\text{Mat}_n\ A$for$n\geq 1$, is coarser than the relation$\sim_{\text{Mor}}$of Morita equivalence. ... 1 The$s_i$are constructed by induction. You've already done the base case ($s_0$). Now assume you've constructed$s_0, \dots, s_{k-1}$and you want to construct$s_k : C_k \to C_{k+1}$such that: $$s_{k-1}d_k - d_{k+1}s_k = id_{C_k} \iff d_{k+1} s_k = s_{k-1}d_k - id_{C_k}.$$ The trick is to see that, while$d_{k+1}$is not surjective, its image is equal to ... 1 Suppose that$F$takes s.e.s. of chain complexes of vector spaces to l.e.s. of vector spaces. Suppose that the objects of$F(0 \to A^{\bullet} \to B^{\bullet} \to C^{\bullet} \to 0)$are$H^0(A)$,$H^0(B)$,$H^0(C)$,$H^1(A)$, etcetera and that the maps$H^i(A) \to H^i(B)$and$H^i(B) \to H^i(C)$are the standard ones. Suppose that$F$commutes with direct ... 1 Let$P$be a nonzero projective in the category. Consider its injective envelope$D$in the category of abelian groups, which is a torsion divisible module. Thus$D$is a direct sum of Prüfer groups and so there is a nonzero morphism$f\colon P\to G$where$G$is a Prüfer group, say$G=\mathbb{Z}(p^\infty)$, for some$p$. Let$G_0=\{0\}\subset G_1\subset ...

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