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Chris Schommer-Pries's user avatar
Chris Schommer-Pries's user avatar
Chris Schommer-Pries's user avatar
Chris Schommer-Pries
  • Member for 15 years, 2 months
  • Last seen this week
  • Notre Dame, IN, United States
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Are all separable algebras Frobenius algebras?
For example what precisely do you mean by a tensor category? Different schools use the term differently. Do you mean just a monoidal category? A rigid monoidal category? Is is $k$-linear? is it Abelian? is the tensor product assumed to be exact? etc, etc.
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Are all separable algebras Frobenius algebras?
I miss-read your question as being about classical algebras in R-modules. You didn't specify what you are assuming about the ground ring and naturally I assumed you were working over a field in which case those extra assumptions are automatically satisfied. Now I see you want to work in an arbitrary tensor category. I wold be surprised if the answer were known in this generality.
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Generalising the parametric transversality theorem to a foliation
If I remember correctly the statement should have the assumption that the adjoint map $M \times S \to N$ is transverse to $R$, where $S$ is the manifold parametrizing the family. ... or something like this. I'll try to look it up later if the OP doesn't do it first.
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Adding inverses to a symmetric monoidal category (Reference?)
Another question... How do you tensor morphisms?
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Adding inverses to a symmetric monoidal category (Reference?)
I am not saying your construction agrees with Quillen's, but that it is similar. The whole point of Thomason's paper is that despite how it looks at first Quillen's construction in fact does not have the analogous universal property (and so any constructions using it to, say, construct multiplications in algebraic K-theory, are bogus). The problem is that certain morphisms, which look like natural transformations, are not actually natural. Your construction looks similar, and so that raises a red flag.
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Adding inverses to a symmetric monoidal category (Reference?)
I am worried about your universal property. How does your construction avoid the problems with Quillen's $S^{-1}S$-construction? (See Thomason "Beware the phony multiplication on Quillen's $S^{-1}S$")
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A tensor product for triangulated categories?
I agree that it is unlikely, but to me not obviously so, hence the question. Maybe a better question is what is the minimal additional structure that I need in order to recover the homotopy category of the tensor products from the two homotopy categories? For example probably having the structure of a stable derivator is sufficient, but that is a lot of data. Can we get away with less? For example what if I remember the action by the homotopy category of spectra $ho(Sp)$?
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Reference for a generalization of Γ-spaces to monoidal model categories
Does anyone know a reference which relates this notion of homotopy algebra to the more classical notion using a cofibrant replacement of the operad? Perhaps just in the cartesian case?
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Pull-back of a fibration along a homotopy equivalence and homotopy classes of sections
Yes on the first point. I will edit. For the second point, my point set topology here is a little rusty, but I don't think we need the compactly generated weak Hausdorff assumption here, though many references will make that assumption. The point is that there is a similar weak factorization system on the category of all topological spaces where instead of just Hurewicz cofibrations we use closed Hurewicz cofibrations. Proving this is part of constructing the "Strom model structure". Finally, all of the cofibrations we use above are variations on mapping cylinders and so should be closed.
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Does "$\forall Z(C(X,Z) \cong C(Y,Z))$" imply $X\cong Y$?
What if you try to mix your two constructions? What I mean is consider spaces Z which are like the Sierpinski 2-point space but where you replace the closed point by an indiscrete space B. Then continuous maps from $X$ into $Z$ are in bijection with pairs $(U,f)$ of an open set $U \subseteq X$ and a set-theoretical map $U \to B$. I think you can recover the poset of opens and the cardinality of each open. That is very close to the space itself. Maybe you can actually get the space?
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Does "$\forall Z(C(X,Z) \cong C(Y,Z))$" imply $X\cong Y$?
One way for the isomorphism to be natural would be if it is induced by composition with a map between $X$ and $Y$. If for all $Z$ you have a bijection of sets $C(X, Z) \cong C(Y,Z)$ which is natural in $Z$, then $X$ and $Y$ are homeomorphic. The homeomorphism and its inverse are constructed by taking $X = Z$ and $Y=Z$, respectively, and looking where the identity maps to. This is a completely general (and standard) categorical argument and doesn't need any topological assumptions nor any kind of topology on the mapping sets.
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Does "$\forall Z(C(X,Z) \cong C(Y,Z))$" imply $X\cong Y$?
If your isomorphism is natural in $Z$, then this is just a special case of the Yoneda lemma. (Your natural homeomorphism then gives a bijection of underlying hom sets, which is all that matters).
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