Timeline for Are algebraic structures uniquely identifed by their free objects?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2021 at 0:53 | comment | added | Benjamin Steinberg | Quasivarieties are also closed under ultraproducts. In any event cancelative monoid and all monoids are ISP classes with the same free object. A variety is determined by its free object on a countable generating set because an identity has only finitely many variables. From this it is easy to see that any identity satisfied by the free object on $\omega$ of satisfied by all algebras in the variety | |
| Jun 14, 2021 at 0:43 | comment | added | Peter Languilla | @BenjaminSteinberg If Im right, an ISP class is equivalent to a quasivariety (class closed under isomorphism, substructures and products), HSP class is equivalent to a variety (class closed under homomorphic images, substructures and products). And $Fr_\omega(V)$ is the free structure of the class with a basis of cardinal $\omega$. | |
| Jun 13, 2021 at 22:38 | comment | added | Benjamin Steinberg | @PeterLanguilla can you explain your terminology | |
| Jun 13, 2021 at 21:44 | comment | added | Peter Languilla | @BenjaminSteinberg can you help me with an example of ISP class (V) that is not determined by Frω(V). Also if you can help me with a hint to prove that an HSP class is determined by Frω(V) will be very usefull. | |
| Jan 26, 2014 at 16:55 | comment | added | Benjamin Steinberg | The free left regular band is free in uncountably many quasivarieties of bands. | |
| Jan 26, 2014 at 16:55 | comment | added | Benjamin Steinberg | The free monoid obviously has the desired property to be the free cancellative monoid. | |
| Jan 26, 2014 at 15:43 | comment | added | Denis | thanks this is useful. So in this case how do you show that the free object indeed has the universal property? | |
| Jan 26, 2014 at 2:31 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |