Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 85592

Homotopy theory, homological algebra, algebraic treatments of manifolds.

1 vote

Coboundary of a cup-product

Let me give an answer for $R=\mathbf Z$, the ring of integers, and let us translate to sheaf cohomology. Your long exact sequence comes from the short exact sequence $$ 0 \rightarrow j_! \mathbf Z \ri …
Johannes Huisman's user avatar
7 votes
Accepted

Is $\mathbb{C}^n\setminus V(f)$ homotopy equivalent with a "large ball complement"?

This is more generally true for semialgebraic subsets of $\mathbf R^n$ and follows from the fact that they are conical at infinity (see Bochnak, Coste, Roy: Real algebraic geometry, Corollary 9.3.7, p …
Johannes Huisman's user avatar
1 vote

Conjugation Cells [equivariant cohomology]

As for your first question, the isomorphism $$ \rho\colon H_C^{2k}(D,S)\rightarrow H^{2k}(D,S) $$ follows from the spectral sequence $$ E_2^{pq}=H^q(C,H^p(D,S))\Rightarrow H_C^{p+q}(D,S). $$ Since $H^ …
Johannes Huisman's user avatar
1 vote
Accepted

Invariants in relative cohomology and compact support cohomology of the quotient

In order to understand the isomorphism, I would translate everything to cohomology of sheaves and then use both Grothendieck's spectral sequences that converge to the same equivariant cohomology group …
Johannes Huisman's user avatar
12 votes

Quotient of solid torus by swapping coordinates on boundary

I believe that your quotient space can be seen as the quotient of the $3$-sphere $S^3$ in $\mathbf C^2$ by the action of complex conjugation. The $3$-sphere $S^3$ can be identified with the join $S^1\ …
Johannes Huisman's user avatar
5 votes
Accepted

Inequivalent free $\Bbb Z/n\Bbb Z$-actions on orientable compact bordered surface

Did you try to use the double $T$ of the surface $S$? Any fixed point-free action of $\mathbf Z/n$ on $S$ induces a fixed point-free action on the closed orientable surface $T$. Moreover, the induced …
Johannes Huisman's user avatar
6 votes

Conceptual proof of classification of surfaces?

Using a little bit of real algebraic geometry, there is a conceptual proof at least in the critical case $\chi=-1$, i.e. the case you're talking about explicitly. Indeed, let $S$ be a compact connecte …
Johannes Huisman's user avatar