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 171227

Homotopy theory, homological algebra, algebraic treatments of manifolds.

5 votes

Injectivity of rationalization on spectrum morphisms

My comment, with more details: No, $E = H\mathbb{Q}$ and $F = \Sigma\mathbb{Z}$ gives a counterexample. $H_0(E) = \mathbb{Q}$ and $H_i(E) = 0$ for $i \neq 0$, so the universal coefficient theorem giv …
user171227's user avatar
5 votes
Accepted

Multi-connected sum decomposition of $n$-manifolds

I don't have a reference, but I think it's not too hard to see that $M_1 \#_k M_2 \approx M_1 \# X_k \# M_2$, where $X_k = (S^1 \times S^{n-1})^{\# (k-1)}$. (Connected sum depends on choices of embed …
user171227's user avatar
5 votes

Are finite $G$-spectra idempotent complete?

To complement Oscar's more systematic answer, let me expand my comment about the case $G = \mathbf{Z}/p\mathbf{Z}$ for a prime number $p$, where the answer is no when $\tilde{K}_0(\mathbf{Z}[G]) \neq …
11 votes
Accepted

Oriented cobordism classes represented by rational homology spheres

The necessary condition pointed out by Jens Reinhold is also sufficient: any torsion class $x = [M] \in \Omega^{SO}_d$ admits a representative where $M$ is a rational homology sphere. EDIT: This is Th …
user171227's user avatar