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How does the solenoid structure of $\mathbb{A}/\mathbb{Q}$ lift to $PGL(2, \mathbb{A})/ PGL(2, \mathbb{Q})$?
Before considering $G(\mathbb Q)\backslash G(\mathbb A)$, consider $G(\mathbb Z[\frac12])\backslash G(\mathbb R\times\mathbb Q_2)$. Also, before $G=PGL_2$, consider $G=\mathbb G_a,\mathbb G_m,SL_2$. 
May 24 
answered  Alexander duality for nonmanifolds 
May 4 
awarded  Good Answer 
Apr 28 
comment 
Intuition behind the definition of quantum groups
A quantum group is a new tensor product on the category of representations. I think that something along the lines of the KZ equations give a coordinatefree construction. At first glance, they just give the braiding, but that's a good start. 
Apr 10 
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The image of the Hurewicz map for rational loop spaces
I expect that is a counterexample, but if she didn't specifically point you to it, I doubt that it is the simplest counterexample. 
Apr 10 
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Reference request: linearly independent cycles in a manifold
Is this a topology question or a linear algebra question? There are two ingredients to reduce to linear algebra: that the manifolds give homology and cohomology classes (so that you have a maps from $k^n$ to $H_j$ and $H^j$); and that transverse intersection is cap product. Now you have a linear algebra question about the rank of a bilinear form, specifically that a bilinear form with invertible determinant has full rank. So it detects every element of $k^n$, yet the form factors through $H_j$, so there was no kernel. 
Apr 6 
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Topological $n$manifolds have the homotopy type of $n$dimensional CWcomplexes
Actually, Tom knows from Milnor that a topological manifold has the homotopy type of a CW complex, which gives the comparison theorem. 
Apr 6 
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Topological $n$manifolds have the homotopy type of $n$dimensional CWcomplexes
Wall's theorem requires singular cohomology as input, so you need another theorem, probably also in Bredon, comparing singular cohomology to sheaf cohomology, just using the manifold hypothesis. 
Apr 5 
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Topological $n$manifolds have the homotopy type of $n$dimensional CWcomplexes
That is overkill and probably circular. 
Apr 4 
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The image of the Hurewicz map for rational loop spaces
Jeff knows that it is not finitely generated in the commutative dga case. mathoverflow.net/questions/182437/… 
Apr 3 
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Pseudomanifolds and Poincaré duality
No, that doesn't work. It is necessary that the parameterizing variety have $\chi=0$. 
Mar 31 
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Pseudomanifolds and Poincaré duality
Here's a modification of the algebraic example that feels less "finite." You can interpret the elliptic curve example saying that elliptic curves have a canonical polarization, so the canonical bundle of curves on $M_{1,1}$ has a projective structure, thus there is an associated bundle of affine cones. Same for $M_g$. But there are more interesting complete curves in $M_g$, so those support a bundle of affine cones, so the singularities twist more. 
Mar 28 
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Pseudomanifolds and Poincaré duality
(but if you are only interested in homology with $\mathbb Q$ coefficients, 2,3,4 are fine) 
Mar 28 
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Pseudomanifolds and Poincaré duality
@DavidC maybe this example can be made algebraic. Replace the the hyberbolic automorphism with the order 6 (not 2,3,4) automorphism of the right elliptic curve. And replace the cone on an elliptic curve with an affine cone. The link of its singularity is not the elliptic curve, but the $S^1$ bundle over it. And replace the circle with a complete variety, say, another elliptic curve. But the surgery is hopeless, so twisted coefficients are out. 
Mar 27 
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Pseudomanifolds and Poincaré duality
Also, the third bullet point in David's definition of pseudomanifold is common, but I think it's a silly axiom. It amounts to the normalization being connected. I guess the point is to reduce the number of orientations, just as a connected manifold has at most one. Connectedness hypotheses are usually a bad idea, but this one is also unwieldy. 
Mar 27 
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Pseudomanifolds and Poincaré duality
I continue to doubt that there are any algebraic examples, but this one has a nice dualizing sheaf, so my earlier suggestion was inadequate to eliminate them. 
Mar 27 
answered  Pseudomanifolds and Poincaré duality 
Mar 22 
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Pseudomanifolds and Poincaré duality
My guess is that the semisimplicity of perverse sheaves shows that if the dualizing sheaf of a variety is not a local system (ie, if it is not a homology manifold), then it does not satisfy duality. At least with char zero coefficients: $\mathbb Q$PD $\implies$ $\mathbb Q$homology manifold. . . . Normal: good question. 
Mar 22 
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Pseudomanifolds and Poincaré duality
How much do you care about the complex hypothesis? Here is a noncomplex example: take a manifold of dimension at least 3, embed an interval, and glue the interval to itself by an involution that reverses the endpoints. 
Mar 14 
awarded  Yearling 