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Aug
11
awarded  Yearling
Aug
7
awarded  Enlightened
Aug
7
awarded  Nice Answer
Apr
28
answered Commuting of Hom and Tensor Product functors?
Apr
20
reviewed Approve Why do we divide to remove elements considered equivalent?
Apr
20
answered Why do we divide to remove elements considered equivalent?
Apr
9
comment The definition of the $false$ truth value
I propose we call your singleton "the lonely heart", @IngoBlechschmidt.
Mar
27
comment Honest application of category theory
Absolutely, @MartinBrandenburg! Algebraic topology really does use that homology and homotopy are functors in a way that would be very tedious to spell out without using the concept of functor. I just picked one example that seemed particularly simple, while also proving something very cool and non-obvious.
Mar
27
comment Honest application of category theory
Sure, for the $n$-dimensional disk substitute $H_n$ or $\pi_n$ for $\pi_1$.
Mar
27
answered Honest application of category theory
Mar
26
comment Examples of functors that preserves products but not equalizers, and vice versa.
A homotopy theorist might call the first functor $\pi_{-1}$.
Mar
21
awarded  Civic Duty
Mar
20
comment Are there any non-obvious colimits of finite abelian groups?
A slightly different way to think about this is that the colimit of a diagram of finite Abelian groups is the "finitization" ("finitization" is the partially defined left adjoint of $U$) of the colimit in all Abelian groups if such a finitization exists. And you've shown that divisible groups do have a finitization, namely $0$.
Mar
20
comment Are there any non-obvious colimits of finite abelian groups?
There's no need to apologize for answering your own question.
Mar
20
comment Are there any non-obvious colimits of finite abelian groups?
I would guess the "well-known theorem" really can fail if $\mathcal{C}$ doesn't have coproducts. I hope someone answers your call for a counterexample.
Mar
19
revised Recognize or interpret this involution : $\frac{\prod_{x\neq i}(1-a_xa_j)}{\prod_{y\neq j}(a_j-a_y)}$
added 65 characters in body
Mar
19
comment Are functors that are left-cancellable necessarily injective on morphisms?
Hint: consider the category $\mathcal{I}$ that has two objects $0$ and $1$, and only one nonidentity morphism $0 \to 1$; what is a functor $G : \mathcal{I} \to \mathcal{C}$?
Mar
19
answered Recognize or interpret this involution : $\frac{\prod_{x\neq i}(1-a_xa_j)}{\prod_{y\neq j}(a_j-a_y)}$
Mar
15
comment why natural transformatoins are also called “morphisms” of functors?
That's a good point, @ZhenLin and it might be exactly the sort of thing the OP is looking for. I'd suggest you write an answer giving some familiar examples, like taking C to be the Lawvere theory for groups, D to be Set (and restricting to product preserving functors), etc.
Mar
15
revised why natural transformatoins are also called “morphisms” of functors?
added 1 character in body