bio  website  math.berkeley.edu/~theojf 

location  Berkeley  
age  29  
visits  member for  5 years, 6 months 
seen  6 hours ago  
stats  profile views  15,094 
I am a recent graduate from UC Berkeley. Starting Fall 2013, I am a Boas Assistant Professor at Northwestern University. My research is broadly centered on quantum field theory — my interests include category theory, representation theory, homological algebra, algebraic topology, Poisson geometry, and theoretical physics.
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awarded  Popular Question 
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Are lax functor categories into a cartesian closed 2category cartesian closed?
Well, I misread, and didn't see you wanted cartesian closed categories, rather than just cartesian categories. But I think via a "functor of points" approach the same coordinatefull calculation works. 
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answered  Are lax functor categories into a cartesian closed 2category cartesian closed? 
Apr 21 
awarded  Nice Answer 
Apr 21 
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Example s.t. the unbased loopspace is not $\Omega X \times X$
The end result being that X = figure eight provides the example @Jens is looking for. 
Apr 17 
accepted  Given a map of classifying spaces, can the target be described as a groupoid quotient of the source mod some action of some (co)kernel? 
Apr 15 
awarded  Nice Answer 
Apr 14 
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Given a map of classifying spaces, can the target be described as a groupoid quotient of the source mod some action of some (co)kernel?
Oh, these formulae certainly make sense for (derived?) algebraic stacks. Do you know if they are known to be "correct" in the algebrogeometric world? E.g. I really do want to work with the algebraic groups G,H qua schemes, and not just work with the topological groups of $\mathbb C$points. 
Apr 14 
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Given a map of classifying spaces, can the target be described as a groupoid quotient of the source mod some action of some (co)kernel?
Awesome. I'll have to think a bit to make sure I understand this formula. Is it clear that this double coset groupoid is groupal, so that I can take B of it? Or on the right did you just mean the space corresponding to the groupoid $G // (H \times H)$? 
Apr 14 
asked  Given a map of classifying spaces, can the target be described as a groupoid quotient of the source mod some action of some (co)kernel? 
Apr 7 
comment 
What's the cardinality of a higher category?
By calling this number "Euler characteristic", you also learn what to do (in some cases) when the homotopy groups are infinite or there are infinitely many of them (if you work with $\infty$groupoids aka homotopy types). 
Mar 27 
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In set theory, is there a name for a function which maps the empty set to zero and all the others to one?
As @Tim says, this function is quite important, and shows up a lot. I had no trouble understanding your question. $0^{0^{S}}$ is a horrible name, although a great formula. I would call your function "$\pi_{1}$", since "$k$truncation" in the homotopy type theory sense is "$\pi_k$", the $k$th fundamental groupoid. (For _pointed_ spaces, by "$\pi_k$" you might mean the $k$th fundamental _group_. But for unpointed spaces, $\pi_0 = $ "connected components", and $\pi_1$ should mean the fundamental groupoid.) 
Mar 20 
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Are $(\infty,1)$categories $A_\infty$ categories?
@Dmitri Awesome, I'll take a look. 
Mar 20 
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Commutation of tensor products with inverse limits in a specific case
That's the one. 
Mar 19 
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Commutation of tensor products with inverse limits in a specific case
As user74230 suggested in an answer below, I assume you mean $\otimes = \otimes_R$? Then there is a paper by Goodearl  I am traveling and don't remember a more precise reference  that studies more generally the map $M \otimes_R \prod_i N_i \to \prod_i(M\otimes N_i)$. My memory is that it is always injective when $R$ is Noetherian, but at that level of generality injectivity can fail when $R$ is not Neotherian. Actually, I think the failure is witnessed by modules that are isomorphic to $R^X$ for some $X$. 
Mar 4 
answered  Establishing Duality in Tannakian Categories 
Mar 4 
awarded  Enlightened 
Mar 4 
awarded  Nice Answer 
Mar 2 
awarded  Notable Question 
Feb 24 
awarded  Nice Answer 