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11h
comment A question regarding Yoneda's lemma.
I'm curious about your example though, seems like you misunderstood some part from the exercise.
1d
comment Prove that the two polynomials intersect each other only at a single point
Ok, i see. I parsed that wrong.
1d
comment Prove that the two polynomials intersect each other only at a single point
"What remains to show is that $D^K_1(\theta)−D^K_2(\theta)$ is neither positive nor negative for all $0<\theta<1/2$ given a $K$" Wait, what? calling the difference $f(\theta)$, and $\theta_0$ the point of intersection, then according to your graphs, $f(\theta)>0$ for $0<\theta<\theta_0$, and $f(\theta)<0$ for $1/2>\theta>\theta_0$
Sep
18
comment 2-category in HoTT: chapter 9 from the HoTT book
Questions should be self-contained, so it'd be nice if you'd actually post the exercise you're refering to.
Sep
10
awarded  Curious
Sep
4
comment Proving a group, $G$, is a group action onto some set, $X$
composition has to hold for arbtrary elements $g_1$, $g_2\in A$. You have only checked the case $g_1 = g_2$ ($=g$, in your proof)
Sep
3
comment Right-adjoint to the inverse image functor
@Patrick Way better than leaving the question unanswered. If you know the answer, you should consider writing one yourself
Sep
2
suggested suggested edit on Let $\displaystyle D= \{z: |x|<1\}$ which of the following is correct
Aug
27
comment Definition of quotient category
There is however a method for computing arbitrary quotients in cat, see e.g. generalized congruences - Epimorphisms in $\mathcal{Cat}$ by Marek A. Bednarczyk et al
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.