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Jul
28
comment How do we get a simplicial homology functor?
Beside Zhen Lins embedding into $\mathbf{sSet}$, you could also take the barycentric subdivision of your simplicial complex, which is weakly homotopy equivalent to the original abstract simplicial complex, and has a canonical orientation.
Jul
28
comment are there examples of “category-like” structures where distinct pairs of objects have hom-sets that aren't disjoint?
@bachmanimoff Note, that it's not an unusual POV that the notion of intersections/disjointness only makes sense between subsets. So even though some authors don't mention the disjointness axiom, it doesn't necessarily mean they don't require it. They may just "omit" it because they defined $\operatorname{Hom}(a,b)$ and $\operatorname{Hom}(c,d)$ as different (and hence of course disjoint) sets in the first place.
Jul
24
awarded  Yearling
Jul
16
comment What is the line going through points $(5, 5, 5), (2, 2, 2) \in \mathbb{R^3}$ when mapped it is mapped to a point in the real projective plane?
en.wikipedia.org/wiki/…
May
8
comment Union of sets of sets
the latter one.
Apr
24
comment Given an arrow from a category to a product of categories, can we evaluate it on an object using its unique decomposition?
Ah, ok. I've been a little bit to quick with my response. So the small letters are all right. There's still an error, though. The capital $C$ in the given diagram should be small.
Apr
23
comment Given an arrow from a category to a product of categories, can we evaluate it on an object using its unique decomposition?
Yes, seems like all the $C$ should be capital, as well as all the Deltas (i.e. $\delta_c$ should be $\Delta_C$)
Apr
17
revised Understanding a complex set description
deleted 7 characters in body
Apr
17
answered Understanding a complex set description
Apr
17
comment equivalence in category
Well, even though you're particular example was wrong, given $f\colon A\times B\to C$, you can still define maps $i_0\colon C\to C\times C, c\mapsto (c,0)$ and $\Delta\colon C\to C, c\mapsto (c,c)$. Then there are morphisms $a,b\colon C\times C\to C\times C$, which induce maps $a \colon i_0\circ f\to \Delta\circ f$ and $b\colon \Delta\circ f\to i_0\circ f$ and are isomorphisms on the images of $i\circ f$ and $\Delta\circ f$, but aren't group isomorphism. So to prove that $M(A,B)$ is indeed a category, you'd have to show that $a\circ b\ne \operatorname{id}_{\Delta\circ f}$
Apr
16
comment equivalence in category
Man, this is bothering me... your claim as presented is definitely wrong, simply because the identity in a category is unique. Thus if $\operatorname{id}\colon C\to C$ is an identity for $f\colon A\times B\to C$, a morphism $g\colon C\to C$, which acts as identity on the image of $f$ can't be an identity of $f$ as well, provided $M(A,B)$ as defined is indeed a category. I have a hard time finding a couterexample, though.
Apr
15
comment Is there a category of categories?
@BrunoStonek I don't think the answer gives the question enough credit, since there is a thing called $\mathbf{CAT}$, which some people even call a category. My knowledge about this stuff is not sufficiant enough to give a good answer, though....
Mar
31
answered Integrating a function
Mar
31
revised Showing hom-sets are disjoint in a morphism category
added 192 characters in body
Mar
31
answered Showing hom-sets are disjoint in a morphism category
Mar
26
asked Cofibrantly generated categories, cardinals, and the Thomason model structure on $\mathbf{Cat}$.
Mar
19
awarded  Revival
Mar
17
awarded  Citizen Patrol
Mar
12
answered Total Order between sets.
Mar
11
comment Determination of a connected groupoid by its objects and by a set of automorphisms.
@Bento: My comment says that given a connected groupoid $G$, and chosing a connecting morphism $h\colon x\to x'$ for a pair of objects $x,x'$. Then for any $h'\colon x\to x'$, there is a $f\in G(x,x)$, such that $h'=hf$. If I didn't miss anything, this yields in particular that given any $h\colon x\to x'$, and the automorphismgroup $G(x,x)$, we can construct any other morphsim $h'\in G(x,x')$ by precomposition of $h$ with elemenst of $G(x,x)$. I.e. $G(x,x')$ is determined by the knowledge of $G(x,x)$ and $h$.