Timeline for Algebraic objects and lifts of their represented functors
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 15, 2014 at 14:09 | comment | added | WMycroft | That's precisely what I wanted when I said constructive, thanks very much Mamuka. | |
| Aug 15, 2014 at 1:20 | comment | added | მამუკა ჯიბლაძე | And then one checks that the identities that $\Omega$-algebras must satisfy translate into commutative diagrams that these operations $\omega_A$ will fit in. For that one uses the fact that not only every $\mathcal C(X,A)$ is an $\Omega$-algebra but also every $\mathcal C(f,A):\mathcal C(X,A)\to\mathcal C(Y,A)$ is an $\Omega$-algebra homomorphism for any $f:Y\to X$ in $\mathcal C$. | |
| Aug 15, 2014 at 1:18 | comment | added | მამუკა ჯიბლაძე | Perhaps maximally explicitly: given any $n$-ary operation $\omega$ on $\Omega$-algebras, here is how to equip $A$ with it. Since $\mathcal C(X,A)$ is an $\Omega$-algebra naturally for any $X$, in particular $\mathcal C(A^n,A)$ is. Then further in particular the result of applying $\omega$ to any $n$ elements of the $\Omega$-algebra $\mathcal C(A^n,A)$ is defined. So further in particular $\omega(\pi_1,...,\pi_n)\in\mathcal C(A^n,A)$ is defined, where $\pi_1,...,\pi_n\in\mathcal C(A^n,A)$ are the product projections. This is your $\omega_A:A^n\to A$. | |
| Aug 14, 2014 at 20:13 | comment | added | Dimitri Chikhladze | Mathoverflow is a site for researchers. So indeed, questions which do not relate to original research better be asked at math.stackoverflow. Moving or deleting is done by administrators once they deem it necessary. | |
| Aug 14, 2014 at 20:10 | comment | added | Dimitri Chikhladze | You're wellcomed :) Not sure what do you mean by constructive, but e.g. to get the operation $A\times A \rightarrow A$ you evaluate $h_A\times h_A(A\times A) \rightarrow h_A (A\times A)$ at $(p_1, p_2)$, where $p_1$ and $p_2$ are the two projections $A\times A \rightarrow A$. | |
| Aug 14, 2014 at 19:50 | vote | accept | WMycroft | ||
| Aug 14, 2014 at 19:49 | comment | added | WMycroft | Got it, thanks a lot for your help. Do you have any insight over how easy it would be to do this in a constructive manner? The other direction in the equivalence is very easy to do constructively. I didn't realise math.stackexchange and mathoverflow were different things, and agree with you. Is it possible to move this question? | |
| Aug 14, 2014 at 18:53 | comment | added | Dimitri Chikhladze | Btw, perhaps this question was more appropriate for math.stackexchange | |
| Aug 14, 2014 at 18:49 | comment | added | Dimitri Chikhladze | $h_A\times h_A$ is isomorphic to $h_{A\times A}$. This is because Yoneda preserves products, and pretty much the definition of a product. Then you use Yoneda $Nat(h_{A\times A}, h_A) \cong C(A\times A, A)$. That is the fact that Yoneda is fully faithful | |
| Aug 14, 2014 at 18:36 | comment | added | WMycroft | Thanks Dimitri, I've been trying to prove this myself, as you can see in the edit, but have been running into a little trouble, could you point me in the right direction? Ideally, I'd like a constructive proof for a concrete setting, e.g. for $\Omega$ being algebras over a commutative ring. Is there a more straightforward way to do this then trying to unwind everything the Yoneda lemma tells us? | |
| Aug 14, 2014 at 17:28 | history | answered | Dimitri Chikhladze | CC BY-SA 3.0 |