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(Co)chain complexes, abelian Categories, (pre)sheaves, (co)homology in various (possibly highly generalized) settings, spectra, derived functors, resolutions, spectral sequences, homotopy categories. Chain complexes in an abelian category form the heart of homological algebra.

2 votes

Image of an $F$-acyclic resolution homotopic to a projective resolution?

This is explained on the n-lab here. In particular in Theorem 3.15.
Simon Wadsley's user avatar
13 votes

Can we categorify the equation (1 - t)(1 + t + t^2 + ...) = 1?

I think the following does what you want: Let R be a polynomial ring in one variable x over a field with its usual grading; i.e. x has degree $1$. Next consider the graded complex that is R in degree …
Simon Wadsley's user avatar
2 votes

How to construct pair of adjoint functors from category A to category A_D(category of diagrams)

There are several pairs of adjoint functors of the kind you desire but it isn't clear to me if any (or all) of them will give you enough projectives in $A^D$. For example for each $d$ in $D$ $\iota_d …
Simon Wadsley's user avatar
10 votes

Why should the tensor product of $\mathcal{D}_X$-modules over $\mathcal{O}_X$ be a $\mathcal...

One way to think about this is that $D$ is the universal enveloping algebra $U(R,L)$ of the $(k,R)$ Lie-Rinehart algebra $\mathrm{Der}_k(R,R)$. Whenever one has such an enveloping algebra one may perf …
Simon Wadsley's user avatar
3 votes

Homology of solvable (nilpotent) Lie algebras

To fill out my comment with a partial answer to the question. Note that given any (left) $\mathfrak{g}$-module $V$ one can compute $H_i(\mathfrak{g},V)$ as $\mathrm{Tor}_i^{U(\mathfrak{g})}(\mathbb{C …
Simon Wadsley's user avatar
3 votes

Homology of solvable (nilpotent) Lie algebras

This is an attempt to prove the (refined) conjecture I made in the comments of my previous answer. Let $\mathfrak{g}$ be a f.d soluble Lie algebra over $\mathbb{C}$. Let $\mathfrak{n}$ be its derived …
Simon Wadsley's user avatar
4 votes
Accepted

Dual of a bimodule

Copied from comments as requested. There isn't enough context in the question but if you know that $B$ is a projective left $R$-module and nothing else there is a fair chance that you want $\mathrm{H …
Simon Wadsley's user avatar
2 votes

A ring such that all projectives are stably free but not all projectives are free?

Example 1.2.2 in Chapter 1 of Weibel's book in progress on K-theory http://www.math.rutgers.edu/~weibel/Kbook.html says that $R_2$ in the notation of your question has a stably free module that is not …
Simon Wadsley's user avatar