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Apr
22
comment The classifying space of open covers of a manifold
I'm sure there is a much simpler proof, but you do get a proof by combining Lemma 5.4.5.10 and Remark 5.4.5.2 from Lurie's Higher Algebra.
Apr
22
reviewed Approve Approximation of a strongly measurable function by a sequence of simple functions.
Apr
22
comment Is the classifying space a fully faithful functor?
In Maps of BZ/p to BG, Dwyer and Wilkerson show that the map Hom(G, SU(2))/conjugation --> [BG, BSU(2)] (that's homotopy classes of maps, not of pointed maps, which is why the domain is conjugation classes of group homomorphisms) is not surjective if G is the symmetric group on 3 letters and also that it is not injective if G is cyclic of order 15.
Apr
22
comment $G$ has size $55$ and acts on a set of size $39$
You got it, @Albert. You should answer this question, just so it doesn't clutter up the list of unanswered question.
Apr
22
comment $G$ has size $55$ and acts on a set of size $39$
Yes, @Albert, that's right so far, and pretty close to done, too.
Apr
22
comment $G$ has size $55$ and acts on a set of size $39$
Hint: Decompose the set of size 39 into orbits for the action. What sizes can the orbits be?
Apr
22
comment Is the classifying space a fully faithful functor?
Isn't the other summand in $H^2(G,\mathbb{Z})$ given by $\mathrm{Hom}(H_2(G,\mathbb{Z}), \mathbb{Z})$, which vanishes because for a finite group $G$, $H_2(G, \mathbb{Z})$ is finite too?
Apr
22
comment The classifying space of open covers of a manifold
To use Quillen's Theorem A you need the comma categories to be contractible, not empty or contractible. If empties were allowed, you could use Quillen's Theorem A to prove that the classifying space of any category is equivalent to that of the empty category!
Apr
17
comment Ambiguity in the definition of group objects
This sort of "ambiguity" causes no problems at all! Things like this can happen for the traditional definition of group: someone might define $G \times G$ as a set of Kuratowski pairs (i.e., $(g,h) = \{\{g\},\{g,h\}\}$); someone else might think $G \times G = G^2$ is the set of functions $\{0,1\} \to G$; a third person might prefer "short pairs" $(g,h) = \{g,\{g,h\}\}$ or functions with domain $\{1,2\}$, etc. These hypothetical people are unlikely to even notice they are talking about different sets $G \times G$!
Apr
12
comment Is the category of simple graphs finitely complete?
When I took a graph theory course as an undergraduate both categories were mentioned (but the word "category" was not). For objects, having loops everywhere or nowhere is of course the same; let's say simple graphs are not allowed to have any loops. Then, in one category the morphisms were called homomorphisms in the course I took: functions sending adjacent vertices to adjacent vertices; in the other category the morphisms were called reflexive homomorphisms: functions sending to adjacent vertices to vertices that are either adjacent or equal.
Apr
12
comment Is the category of simple graphs finitely complete?
With the correction, (3) is now correct.
Apr
3
comment Criterion for isomorphism of two groups given by generators and relations
To prove two groups given by presentations are isomorphic you can define homorphisms in both directions and prove both composites are the identity, here's an example.
Apr
3
revised $G=\langle a,b \mid abab^{-1}\rangle$ and $H=\langle c,d \mid c^2d^2\rangle$ are isomorphic (Can't use Seifert/van Kampen Theorem)
added 138 characters in body
Apr
3
comment $G=\langle a,b \mid abab^{-1}\rangle$ and $H=\langle c,d \mid c^2d^2\rangle$ are isomorphic (Can't use Seifert/van Kampen Theorem)
Right, @Craig: to define a homomorphism $f : G \to H$ you just need to assign an element of $H$ to each generator of $G$ and check that the assignment would send every relation in the presentation of $G$ to the identity element of $H$. As you said, if $f$ is bijective then it is an isomorphism, but sometimes that might be hard to show directly. One way is to give another homomorphism the other way, $g:H\to G$ and to check that $f\circ g = id_H$ and $g\circ f=id_G$ (for example, by checking that $f(g(x))=x$ for each generator in the presentation of $H$ and the same for the other composite).
Apr
3
answered $G=\langle a,b \mid abab^{-1}\rangle$ and $H=\langle c,d \mid c^2d^2\rangle$ are isomorphic (Can't use Seifert/van Kampen Theorem)
Mar
28
revised Proof that the tensor product is the coproduct in the category of R-algebras
corrected a few typos (that someone else found and fixed, but their proposed edit was rejected for some reason)
Mar
15
comment Ring structure with underlying abelian group $G$
I think the standard example of a ring with left identities but no right identity is $\left\{ \pmatrix{x&y\\x&y} : x,y \in \mathbb{R} \right\}$.
Mar
14
comment Strongly unbiased symmetric monoidal category
I'm not sure that for Question 2 you really need to use the coherence theorem; it feels more like proving that the category of finite sets is equivalent to it's skeleton. Pick for each finite set $I$ a bijection $\beta_I : I \to \{1, 2, \cdots, |I|\}$ so that $\beta_{\{1,2,\cdots,n\}}$ is the identity and use those $\beta_I$'s to construct a strongly unbiased symmetric monoidal category given an unbiased one. I haven't thought it through but I wouldn't expect you need coherence to prove this is an equivalence (I could easily be wrong).
Mar
14
comment Strongly unbiased symmetric monoidal category
We'll see if any answers turn up, but I wouldn't be surprised at all if this were unwritten folklore.
Mar
14
answered Help with proof: given outer measure $m^*$, if $A = \bigcup_{n =1}^\infty A_n$, then $m^*A \leq \sum\limits_{n =1}^\infty m^*(A_n)$