Search Results

It is not true in general. Let $n=3$, and take a piece of a brick wall (thickness one brick) that is homeomorphic to a 3-disc. For example, just take a largish rectangular wall. There are bricks of …

One reason is that it is easier to make equivariant open coverings of $G$-spaces than to make equivariant triangulations of $G$-spaces. (I was going to add this as a comment but saw the script warning …

As in Tom Goodwillie's comment, if you take the construction that you discuss for a $G$-set $X$, the equivariant homotopy type of the space obtained before you quotient out by the action of $G$ is cal …

I needed a slightly more complicated example than this in a paper of mine, so I included a proof there. For any non-trivial finite group $Q$ I give a 3-dimensional rationally acyclic complex with a …

Here is a partial answer. There are finite acyclic 2-complexes that are not contractible. Any such 2-complex can be embedded in $\mathbb{R}^5$, and a regular neighbourhood of such a 2-complex will b …

Corollary 5.4.2 of Wall's article `Poincaré complexes I', Ann. Math. 86 (1967) 213-245 gives examples of 4-dimensional Poincaré complexes $X$ with fundamental group of prime order $p\geq 23$ for which …

The $E_1$ page does not tell you what the higher differentials will be, and you will have to know at least something about the integral cohomology. Consider the case when $X$ is a Moore space $M(1,\m …

Yes, there is such a decomposition. The spheres on which $\mathbb{Z}/p\mathbb{Z}$ acts freely are necessarily odd dimensional for $p>2$. View each $S^{2m-1}$ as the unit sphere in $\mathbb{C}^m$, wi …

Take a surface (without boundary but not compact) of infinite genus, and tesselate it by polygons so that there are no non-trivial combinatorial symmetries of the surface. The universal covering of …

Let $G=(\mathbb{Q},+)$. Then ${\rm cd}(G)=2$ and ${\rm cd}(G\times G)=3$.

There has been a lot of work on cases of this conjecture connected to Coxeter groups. M. Davis and R. Charney made a conjecture that comes from these cases in 1995 in The Euler characteristic of a no …

The Kan-Thurston construction depends not just on the homotopy type of $X$, but it depends very heavily on the exact choice of cell structure for $X$. However, it does have some naturality properties …

Your quotient manifold is homeomorphic to $S^3$. I know this because it is a closed 3-manifold and its fundamental group is trivial, so I'm quoting the Poincare conjecture/Perelmann's theorem. The f …

The mod-2 homology groups of any closed $n$-manifold are `$n$-palindromic' by Poincare duality, i.e., $H_m(M;\mathbb{Z}/2)\cong H_{n-m}(M;\mathbb{Z}/2)$ for each $m$. If the mod-2 homology of $X$ is …

Here is model that is obviously functorial: take for $\underline{E}G$ the simplicial complex with vertex set the finite subsets of $G$ and simplices the finite chains of sets ordered by inclusion. A …