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Mar
15 |
awarded | Popular Question |
Feb
20 |
awarded | Enlightened |
Feb
20 |
awarded | Nice Answer |
Jul
30 |
comment |
Does every Lawvere theory arise in this way?
@goblin is perfect. |
Jul
28 |
comment |
Does every Lawvere theory arise in this way?
Can you give a more detail on what $Lawv(X)$ should be? It should be the smallest finite product sub-category of $\mathbf C$ containing $X$? It should be the smallest finite product category generated by $X$ and its algebraic morphism? |
May
9 |
awarded | Yearling |
Apr
29 |
awarded | Good Answer |
Feb
20 |
answered | The groupoid of algebraic expressions and proofs |
Feb
17 |
awarded | Favorite Question |
Dec
5 |
answered | Relations between ordinary functor categories and higher categories |
Dec
2 |
awarded | Critic |
Nov
9 |
awarded | Necromancer |
Jul
2 |
answered | Basic category theory: Universality of adjunction unit is justified by Yoneda Proposition in Mac Lane's text |
Jul
2 |
comment |
Basic category theory: Universality of adjunction unit is justified by Yoneda Proposition in Mac Lane's text
I don't understand, what you mean by "of course, this is not explicitly defined in general, since it depends on the particular adjunction"? |
Jul
2 |
awarded | Curious |
May
21 |
comment |
Simple show cases for the Yoneda lemma
@HansStricker what do you mean by $K_1$ and $K_2$ are the representable functors? |
May
21 |
comment |
A categorical characterization of ordinal numbers
I guess that for help could be useful to characterize $\star$ by universal property, otherwise is not clear what a category (not necessarily a subcategory of $\mathbf{Cat}$) closed by $\star$ should be. |
May
20 |
comment |
A categorical characterization of ordinal numbers
Note that in the note you have linked Joyal define $\star B$ as a functor of type $\mathbf {Cat} \to B\setminus \mathbf {Cat}$ not $\mathbf {Cat} \to \mathbf {Cat}$. |
May
20 |
comment |
A categorical characterization of ordinal numbers
The operation $\star$ does not commute with colimit: indeed if that where true then for every limit ordinal $\lambda=\bigcup_{\gamma < \lambda} \gamma$ you should have that $\lambda+1=\lambda\star[0]=\bigcup_{\gamma < \lambda} \gamma\star[0]=\bigcup_{\gamma < \lambda}\gamma+1=\lambda$ things that's not possible since $\lambda$ doesn't have a max while $\lambda+1$ does. |
May
9 |
revised |
Further relation between monads and theories
fixed grammar |