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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
$$
0 \rightarrow j_! \mathbf Z \ri …
7
votes
Accepted
Is $\mathbb{C}^n\setminus V(f)$ homotopy equivalent with a "large ball complement"?
This is more generally true for semialgebraic subsets of $\mathbf R^n$ and follows from the fact that they are conical at infinity (see Bochnak, Coste, Roy: Real algebraic geometry, Corollary 9.3.7, p …
1
vote
Conjugation Cells [equivariant cohomology]
As for your first question, the isomorphism
$$
\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^ …
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 …
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\ …
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 …
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 …