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Topology of cell complexes and manifolds, classification of manifolds (e.g. smoothing, surgery), low dimensional topology (e.g. knot theory, invariants of 4-manifolds), embedding theory, combinatorial and PL topology, geometric group theory, infinite dimensional topology (e.g. Hilbert cube manifolds, theory of retracts).
7
votes
Accepted
What are the covering spaces of $\mathbb{Q}$?
Yes.
For a cardinal $c$ let $V_c\subset X$ be the set of all $x$ such that $c$ is the cardinality of $p^{-1}(x)$. So $X$ is the union of some set of disjoint nonempty open sets $V_c$ such that when re …
2
votes
Accepted
Preserving simple-connectedness under intersection complexes
Yes.
There is a map $X\to X_U$, taking each point $x$ to some point in the simplex corresponding to the set of all $i$ such that $x\in U_i$.
This map induces a surjection $\pi_1(X)\to \pi_1(X_U)$. To …
4
votes
Accepted
Borel cohomology for circle actions on odd spheres
The action of $S^1$ on $\mathbb C^k$ yields a rank $k$ complex vector bundle $V=ES^1\times_{S^1}\mathbb C^k$ on $BS^1$. You are interested in the cohomology of $S(V)=ES^1\times_{S^1}S^{2k-1}$, the tot …
3
votes
Accepted
Understanding $(\mathbb{Z}/3)^2 \times_{\mathbb{Z}/3} M$
This has been more or less said in the comments, but it seems like someone should write an answer.
If $H$ is a subgroup of $G$ and $H$ has a left action on a space $X$, then a space $G\times _HX$ is d …
17
votes
Accepted
Can a cyclic group of prime order act on a rationally acyclic finite dimensional complex and...
Yes, I think you can make an example like this (for $p=2$, but it generalizes).
Let $R$ be the group ring $\mathbb Z[C_2]=\mathbb Z[x]/(x^2-1)$. Make a chain complex of free $R$-modules
$$
M_0 \leftar …
10
votes
Accepted
Do $h$-cobordism groups arise from a 'Thom-like' spectrum?
This topic has a very different flavor from what is usually meant by cobordism. No, the correspondence between cobordism classes and homotopy classes (of a Thom space or Thom spectrum) has no analogue …
22
votes
Fixed-point free diffeomorphisms of surfaces fixing no homology classes
Here's an example with $g=2$. Let $T$ be the torus $\mathbb C/L$, where $L$ is the lattice spanned by $1$ and $\zeta=e^{2\pi i/6}$. Let $f:T\to T$ be induced by multiplication by $\zeta$. This is a di …
4
votes
Accepted
On the proof of the surgery step in Wall's book
The theorem has the hypothesis "$f$ is in this class", meaning that the embedding $f$ is in the regular homotopy class of immersions determined by $F$ together with the element of the relative homotop …
8
votes
Contractible set in a manifold
No. Let $M$ be $S^1\times S^2$ and let $K$ be homeomorphic to $S^1$ and chosen such that in the universal cover $\tilde M=\mathbb R\times S^2$ there are two liftings $K_1$ and $K_2$ of $K$ which are l …
11
votes
Accepted
The (co)tangent sheaf of a topological space
Your $𝑇𝑋$ is always $0$. If $𝐷$ is a derivation and $𝑓$ is a function, then for every point $𝑥$ $𝐷𝑓$ vanishes at $𝑥$; it suffices to prove this when $𝑓(𝑥)=0$, and in that case $𝑓=𝑔ℎ$ wher …
2
votes
Extending a continuous map over projective space
For larger $n$ we can still make a counterexample, can't we?
Let $S$ be a set of $n+1$ points in general position, and choose them to have rational coordinates. Any linear map which fixes these poin …
9
votes
Accepted
Is the boundary of a manifold topologically unique?
Since you say in a comment that you might be satisfied with a homotopy equivalence, let me sketch a proof that the homotopy type of the boundary depends only on the interior.
Let $Y$ be the interior …
2
votes
Accepted
Isotopy extension theorem: how non-unique is ambient isotopy
If I interpret the question correctly then the answer is "yes". You seem to be asking whether, if $H'$ is an isotopy satisfying the same conditions as $H$, there must be a one-parameter family of such …
5
votes
$S^1$-quotient of the space of unbased contractible loops of a finite dimensional $K(\pi,1)$
THIS ANSWER IS WRONG (in that it claims the result in excess generality)
Suppose that $X$ is a $K(\pi,1)$ and a CW complex. Let $E$ be a universal covering space of $X$, so that $E$ is contractible a …
12
votes
Accepted
Are open orientable 3-manifolds parallelizable via obstruction theory?
I think you are saying: for a closed $3$-manifold, vanishing $w_1$ implies vanishing $w_2$ by Wu's relations, but is this still true if the manifold is not closed? The answer is yes.
For a compact ma …