# Tag Info

9

$H_1(S^1)$ is the kernel of the map $\mathbb{Z}^2\to\mathbb{Z}$. Every such homomorphism is of the form $(x,y)\mapsto ax+by$ for some $a,b\in\mathbb{Z}$. In this case the map is surjective, so $\gcd(a,b)=1$, and the kernel is the infinite cyclic group generated by $(b,-a)$.

6

If $R$ is any ring, then the forgetful functor $\mathsf{Mod}(R) \to \mathsf{Ab}$ is exact. Remember that exact := left exact + right exact := preservation of finite limits and finite colimits. In fact, the forgetful functor preserves all limits and colimits. The reason is that one really constructs limits and colimits of $R$-modules as the limit and colimits ...

4

There are Chern classes defined on K-theory groups with values in cyclic homology and Hochschild homology, which are the exact analogue of characteristic classes for general algebras —in fact, these can be defined using those. You can find this explained in Loday's book on cyclic homology, in Karoubi's book on the same subject, in Rosenberg's introduction ...

4

$\mathscr{T}/\mathscr{C}$ still makes sense if $\mathscr{C}$ is not strict, but it's exactly the same as taking the quotient by the "strictification" of $\mathscr{C}$ (i.e., the subcategory consisting of all objects isomorphic to an object of $\mathscr{C}$), so you don't get anything more general.

3

A typical example is the category of vector bundles on a (non-pathological) topological space. The corresponding algebraic example is the category of finitely generated projective modules over some (non-pathological) commutative ring. Notice: If $K \to P$ is a kernel of $P \to Q$ in the category of f.g. proj. $R$-modules, then this is also the kernel in ...

3

First, $\text{gl. dim}(R) < \infty$ does not imply $\text{gl. dim}(R/x) < \infty$ for any regular $x \in \mathfrak{m}$ - indeed, this would mean that all complete intersections are regular. What is true is that for $R$ regular local, $R/x$ is regular iff $x \in \mathfrak{m} \setminus \mathfrak{m}^2$. For the proof of $\text{gl. dim}(R) < \infty ... 3 You can use Hilbert Syzygy theorem to state that the projective dimension of$M$is finite, and is at most$r$. Now you can use Auslander Buchsbaum formula (it works also with graded$\mathbb K[x_1,\ldots, x_r]$modules): $$pd(M)=depth(S)-depth(M)=r-depth(M).$$ Now the fact that there exist an homogeneous regular element ensure that$depth(M)$is at least ... 2 If$A$is divisible and$A \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$vanishes, then$A \to \mathbb{Q}$vanishes. In fact,$A \to \mathbb{Q}$factors through$A \to \mathbb{Z}$by assumption, so that the image is a divisible subgroup of$\mathbb{Z}$. The only divisible subgroup of$\mathbb{Z}$is trivial. This shows that$\mathbb{Q} \to ...

2

She could probably define singular (co)homology and do a couple of applications in two hours, if she just stated the main technical theorems. Possibly axiomatic homology theory would be a beginning point suited to the abstract and algebraic pitch of the course. Something more interesting would involve spectra and triangulated categories, but that's surely ...

2

Since the $C_i$ are free, only the ranks matter, and we may tensor with $\mathbb{Q}$. Then the complex becomes exact because the homology was assumed to be torsion. Now use that the Euler characteristic of an exact complex vanishes.

2

hint: you could replace the sequence by $Z^2 \to B \to ker(Z^n \to Z) \to 0$

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I think that if Vopěnka's principle is true, then $\mathsf{Mod}(R)^{op}$ can't fully embed in $\mathsf{Mod}(S)$ for any non-zero $R$ and $S$ (all the references that follow are to "Locally Presentable and Accessible Categories" by Adámek and Rosický): If it did, then $\mathsf{Mod}(R)^{op}$ would be bounded (Theorem 6.6), and since it is also complete it ...

1

For arbitrary bimodules $P$ and $Q$ you get a Morita context with both $f=0$ and $g=0$, and you can certainly choose the bimodules to be progenerators for $A$ and $B$ respectively. For the "well known fact" that you mention, I think you also need that $$f:P\otimes_BQ=P\otimes_B\operatorname{Hom}_A(P,A)\to A$$ is the natural map (i.e., the evaluation map ...

1

We have the exact sequences $$(1) \quad 0 \to \ker f \to C \to \mathrm{im }f \to 0$$ and $$(2) \quad 0 \to \mathrm{im }f \to D \to \mathrm{coker }f \to 0.$$ The long exact sequence of homology coming from $(1)$, along with the assumption that $\ker f$ is acyclic, shows that $C \to \mathrm{im} f$ is a quasi-isomorphism. The long exact sequence of ...

1

I once had a compendium of Jan-Erik Roos in which derived functors was compared with derivatives: derivatives measured the deviation from being a straight line, while a derived functor (as I remember the Ext-functor) was measuring the deviation of a module from being a vector space. Of course this was very informally and I have never seen anything like that ...

1

Here is a suggestion: identities among relations for groups and van Kampen diagrams. One source is this book. Also van Kampen diagrams occur in many books on group theory. The topic allows lots of pictures and relates to group theory. Part of the motivation is "how to specify infinite objects, such as infinite groups?". Also how do you find ...

1

Probably the definition of "principal ideal ring" requires that $\Gamma$ is a domain.

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You could tensor it with $\mathbb{Q}$ (or any field of characteristic 0) to show that $H_1(S^1)$ is isomorphic to $\mathbb{Z} \oplus F$ for some finite abelian group $F$. The reason for this is that you end up with the short exact sequence of vector spaces $$0 \to H_1(S^1) \otimes \mathbb{Q} \to \mathbb{Q}^2 \to \mathbb{Q} \to 0$$ and so dimension ...

1

HINT: Imbed $A$ into an acyclic object $B$ and consider the short exact sequence $$0\to A \to B \to C \to 0$$ Use the long exact sequence involving the derived functors of $F$. You get $F^1(A)=0$. The higher derived functors do not have to be zero. Take for instance $Ext(M, \cdot )$ as a left exact functor. $A$ has the exactness property $\iff$ ...

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