38,452 reputation
244116
bio website
location
age
visits member for 3 years, 10 months
seen 15 mins ago

Oct
7
comment Single $\Delta$-complex structure on $S^3$
Well, you can take $\Delta^3$ and collapse the boundary to a point. But that's not a $\Delta$-complex.
Oct
7
comment Axioms of Abelian Category
You should spell out what (2) and (3) are. It is annoying to have to load a 764-page PDF to find two sentences.
Oct
6
comment Two different definitions of sheaf of $K$-modules and tensor products.
If your space is locally connected then there is a straightforward argument showing that the two coincide, following the outline you suggest. But it should be true in general.
Oct
6
comment Injective resolution of complexes equivalent to regular definition
Unless your definition of "complex" excludes negative degrees, this is an extra condition in general.
Oct
5
comment De Rham Cohomology is Sheaf Cohomology
Yes. You don't mean to say it's trivial, do you?
Oct
5
comment De Rham Cohomology is Sheaf Cohomology
Your argument is correct. However, it only shows that you have an exact sequence – still a long way from calculating sheaf cohomology.
Oct
5
comment Is there accepted terminology for algebraic structures whose every subalgebra is free?
One might say "hereditarily free".
Oct
4
comment Category theory? Logic? Anyone experienced this like me?
Having a transitive model is not as convenient. When category theorists adopt a universe axiom it is always for convenience...
Oct
3
comment Elementary motivations for free resolutions
One might ask the same questions about higher Ext or Tor groups...
Oct
3
comment Weighted colimits,hom-functor,Usage of Yoneda lemma
No, I refuse. That step is very basic and you must be able to work out at least that much for yourself if you want to get anywhere with category theory.
Oct
3
comment Weighted colimits,hom-functor,Usage of Yoneda lemma
Strictly speaking one uses the Yoneda lemma as well as the uniqueness of representing objects for representable functors.
Oct
3
comment Weighted colimits,hom-functor,Usage of Yoneda lemma
These are easy. Do you really understand what I'm writing or not?
Oct
3
comment Weighted colimits,hom-functor,Usage of Yoneda lemma
There is nothing wrong with the syntax. $\mathbf{Set}(D, X)$ is a functor of type $\mathcal{E}^\mathrm{op} \to \mathbf{Set}$, of course.
Oct
3
comment Weighted colimits,hom-functor,Usage of Yoneda lemma
There's no use in me doing the calculation if you don't understand it. Do you know the calculus of ends or not?
Oct
3
comment Weighted colimits,hom-functor,Usage of Yoneda lemma
Substitution is just substitution. $E(d, -)$ is a diagram $E \to \mathbf{Set}$, is it not? The tricky bit is applying the Yoneda lemma. I would use the calculus of ends here...
Oct
3
comment If I want to avoid quantifiers?
It is true that you can replace $\forall x \in X . P (x)$ with $\forall x . (x \in X \to P (x))$. But you can't get rid of $\forall$ entirely.
Oct
3
comment If I want to avoid quantifiers?
Then what do you mean by your first paragraph?
Oct
3
comment If I want to avoid quantifiers?
No. There is a world of difference between $\forall x . (P (x) \to Q)$ and $(\forall x . P (x)) \to Q$, and if you omit $\forall$ it is not possible to distinguish between the two.
Oct
2
comment Equivalence of categories $\Pi_1:\mathbf{K^1_{CW,*}}\to \mathbf{Grp}$
What? The higher homotopy groups $\pi_n$ have a perfectly good classical definition not involving any $n$-groupoids.
Oct
2
comment Equivalent Conditions for left-adjoint-left-inverse
Just do it by hand. It's the usual transport-of-structure argument. (For example, if $G$ is a group and $X$ is a set such that $X \cong G$, then there is a group structure on $X$ making the given bijection a group isomorphism.)