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Homotopy theory, homological algebra, algebraic treatments of manifolds.

7 votes

Mayer-Vietoris implies Excision

I'd have taken "excision" to mean that the map $H_n(A,A\cap B)\rightarrow H_n(X,B)$ is an isomorphism. I'd have taken "Mayer-Vietoris" to mean not just the exactness of some sequence $$\cdots\rightar …
Steven Landsburg's user avatar
34 votes
Accepted

Coefficients in cohomology

This (elementary and perfectly standard) example might help show the power of sheaves with non-constant coefficients: First, think about the circle $S^1$. Suppose you want to understand (real) line …
Steven Landsburg's user avatar
13 votes

Which of Quillen's Papers Should I read?

Quillen's book on Homotopical Algebra is a great pleasure to read, and likely to appeal to a geometer.
8 votes

Are there 'cohomology' functors that respect all Eilenberg-Steenrod axioms except homotopy i...

For a manifold $X$, define $H_0(X)$ to be the direct sum of all the tangent spaces to $X$. This extends in the obvious way to a functor on the category of manifolds and smooth maps. For a pair $(X,A …
Steven Landsburg's user avatar
1 vote

On the simply connectedness of Symmetric products and Hilbert schemes of points

Re your first question: If $\pi_1(X)=H_1(X,{\mathbb Z})=0$, then any $m$-th symmetric product of $X$ is simply connected; this is a special case of Theorem 1.1 in this paper by Kallel and Taamallah.
Steven Landsburg's user avatar
11 votes

Quillen's motivation of higher algebraic K-theory

As an appendix to Tom Harris's nice answer, it might be worth mentioning that the idea of defining algebraic K-groups as the homotopy groups of something was certainly in the air before Quillen, e.g. …
Steven Landsburg's user avatar
3 votes

Picard group, Fundamental group, and deformation

First: The identity map on $X$ factors through the inclusion from $X$ to $X\times {\mathbb A}^1$. Therefore $Pic(X)$ is a direct summand of $Pic(X\times{\mathbb A}^1)$. This is the same argument yo …
Steven Landsburg's user avatar
7 votes
Accepted

Homotopy invariance of $K_0$

I will use capital letters to denote $R[t]$ modules and capital letters subscripted with $0$ to denote $R$ modules. All rings are assumed noetherian and all modules are assumed finitely generated. T …
Steven Landsburg's user avatar
21 votes
Accepted

Other Homology Theories still Count Holes?

If your homology theory is of the form $H_n(X) = H_n(S(X)) $ where $S$ is some functor from your original category to non-negative chain complexes, then the Dold-Kan correspondence gives you a corr …
Steven Landsburg's user avatar
8 votes

What are some interesting problems in the intersection of Algebraic Number Theory and Algebr...

Check out Vic Snaith's work on Explicit Brauer Induction.
11 votes
0 answers
305 views

Generalized Classical Adjoints and Factorizations of the Characteristic Polynomial

This is idle noodling, and I'm prepared to learn that it's foolish as well as idle. But.... Let $M$ be an $n\times n$ matrix over, oh, let's say an algebraically closed field for now. There have be …
Steven Landsburg's user avatar
2 votes

A toolbox for algebraic topology

Re your second question, I don't have it in front of me but I believe you'll find this in Artin and Mazur's book "Etale Homotopy Theory", near the very beginning of the first section.
1 vote

Lefschetz fixed point formula: an 'easy' proof; cohomology with compact support

Kleiman's paper in the "Dix Exposes sur la cohomologie des schemas" volume might be what you're looking for.
Steven Landsburg's user avatar
17 votes
Accepted

Algebraic K-theory and Homotopy Sheaves

For the Zariski topology, one has cohomological descent if $R$ is regular. (This yields the Brown-Gersten spectral sequence.) For the etale topology, still assuming $R$ regular, descent fails for …
Steven Landsburg's user avatar
4 votes

Hypercohomology of a complex of sheaves that might be acyclic (or might not)

I'm not sure I completely understand your question, but: There is a double complex with $\Gamma(C^pF^q)$ in the $(p,q)$ place. By taking cohomology first vertically and then horizontally, or vice ve …
Steven Landsburg's user avatar

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