Pushout of unital non commutative algebras











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I like to know if there is a pushout in the category of non commutative alegbras with unit and if the answer is "yes", who is it?










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    Yes. It's a variant of the amalgamated free product, adapted to algebras.
    – Qiaochu Yuan
    Nov 17 at 23:49










  • @QiaochuYuan, do you have some reference? I would like to check the construction carefully and if your reference has it with $ast$-algebras, it would be better. I'd really appreciate that
    – GaSa
    Nov 18 at 1:35






  • 1




    You might not find a reference that's amenable to careful checking. Careful proofs of a theorem like this are more likely to be done in much more generality-for arbitrary "categories of algebras," for instance.
    – Kevin Carlson
    Nov 18 at 1:39










  • ok @KevinCarlson. Well, it would be ok if I just see who is this algebra and how I can define the morphisms involved in the diagrams.
    – GaSa
    Nov 18 at 1:54















up vote
1
down vote

favorite
1












I like to know if there is a pushout in the category of non commutative alegbras with unit and if the answer is "yes", who is it?










share|cite|improve this question


















  • 1




    Yes. It's a variant of the amalgamated free product, adapted to algebras.
    – Qiaochu Yuan
    Nov 17 at 23:49










  • @QiaochuYuan, do you have some reference? I would like to check the construction carefully and if your reference has it with $ast$-algebras, it would be better. I'd really appreciate that
    – GaSa
    Nov 18 at 1:35






  • 1




    You might not find a reference that's amenable to careful checking. Careful proofs of a theorem like this are more likely to be done in much more generality-for arbitrary "categories of algebras," for instance.
    – Kevin Carlson
    Nov 18 at 1:39










  • ok @KevinCarlson. Well, it would be ok if I just see who is this algebra and how I can define the morphisms involved in the diagrams.
    – GaSa
    Nov 18 at 1:54













up vote
1
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up vote
1
down vote

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1





I like to know if there is a pushout in the category of non commutative alegbras with unit and if the answer is "yes", who is it?










share|cite|improve this question













I like to know if there is a pushout in the category of non commutative alegbras with unit and if the answer is "yes", who is it?







category-theory noncommutative-algebra






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asked Nov 17 at 20:55









GaSa

536




536








  • 1




    Yes. It's a variant of the amalgamated free product, adapted to algebras.
    – Qiaochu Yuan
    Nov 17 at 23:49










  • @QiaochuYuan, do you have some reference? I would like to check the construction carefully and if your reference has it with $ast$-algebras, it would be better. I'd really appreciate that
    – GaSa
    Nov 18 at 1:35






  • 1




    You might not find a reference that's amenable to careful checking. Careful proofs of a theorem like this are more likely to be done in much more generality-for arbitrary "categories of algebras," for instance.
    – Kevin Carlson
    Nov 18 at 1:39










  • ok @KevinCarlson. Well, it would be ok if I just see who is this algebra and how I can define the morphisms involved in the diagrams.
    – GaSa
    Nov 18 at 1:54














  • 1




    Yes. It's a variant of the amalgamated free product, adapted to algebras.
    – Qiaochu Yuan
    Nov 17 at 23:49










  • @QiaochuYuan, do you have some reference? I would like to check the construction carefully and if your reference has it with $ast$-algebras, it would be better. I'd really appreciate that
    – GaSa
    Nov 18 at 1:35






  • 1




    You might not find a reference that's amenable to careful checking. Careful proofs of a theorem like this are more likely to be done in much more generality-for arbitrary "categories of algebras," for instance.
    – Kevin Carlson
    Nov 18 at 1:39










  • ok @KevinCarlson. Well, it would be ok if I just see who is this algebra and how I can define the morphisms involved in the diagrams.
    – GaSa
    Nov 18 at 1:54








1




1




Yes. It's a variant of the amalgamated free product, adapted to algebras.
– Qiaochu Yuan
Nov 17 at 23:49




Yes. It's a variant of the amalgamated free product, adapted to algebras.
– Qiaochu Yuan
Nov 17 at 23:49












@QiaochuYuan, do you have some reference? I would like to check the construction carefully and if your reference has it with $ast$-algebras, it would be better. I'd really appreciate that
– GaSa
Nov 18 at 1:35




@QiaochuYuan, do you have some reference? I would like to check the construction carefully and if your reference has it with $ast$-algebras, it would be better. I'd really appreciate that
– GaSa
Nov 18 at 1:35




1




1




You might not find a reference that's amenable to careful checking. Careful proofs of a theorem like this are more likely to be done in much more generality-for arbitrary "categories of algebras," for instance.
– Kevin Carlson
Nov 18 at 1:39




You might not find a reference that's amenable to careful checking. Careful proofs of a theorem like this are more likely to be done in much more generality-for arbitrary "categories of algebras," for instance.
– Kevin Carlson
Nov 18 at 1:39












ok @KevinCarlson. Well, it would be ok if I just see who is this algebra and how I can define the morphisms involved in the diagrams.
– GaSa
Nov 18 at 1:54




ok @KevinCarlson. Well, it would be ok if I just see who is this algebra and how I can define the morphisms involved in the diagrams.
– GaSa
Nov 18 at 1:54










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Given unital $R$-algebras $Aleftarrow Bto C$, the pushout $A star_B C$ is generated as an $R$-algebra by generators of $A$ and of $C$, modulo the union of the relations in $A$ and in $C$, as well as further relations identifying the two resulting images of each element of $B$. This immediately gives the canonical maps from $A$ and $C$. You can describe this construction as a quotient of the free $R$-module on words with letters from $A$ and $C$, if you like.






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    Given unital $R$-algebras $Aleftarrow Bto C$, the pushout $A star_B C$ is generated as an $R$-algebra by generators of $A$ and of $C$, modulo the union of the relations in $A$ and in $C$, as well as further relations identifying the two resulting images of each element of $B$. This immediately gives the canonical maps from $A$ and $C$. You can describe this construction as a quotient of the free $R$-module on words with letters from $A$ and $C$, if you like.






    share|cite|improve this answer

























      up vote
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      Given unital $R$-algebras $Aleftarrow Bto C$, the pushout $A star_B C$ is generated as an $R$-algebra by generators of $A$ and of $C$, modulo the union of the relations in $A$ and in $C$, as well as further relations identifying the two resulting images of each element of $B$. This immediately gives the canonical maps from $A$ and $C$. You can describe this construction as a quotient of the free $R$-module on words with letters from $A$ and $C$, if you like.






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        Given unital $R$-algebras $Aleftarrow Bto C$, the pushout $A star_B C$ is generated as an $R$-algebra by generators of $A$ and of $C$, modulo the union of the relations in $A$ and in $C$, as well as further relations identifying the two resulting images of each element of $B$. This immediately gives the canonical maps from $A$ and $C$. You can describe this construction as a quotient of the free $R$-module on words with letters from $A$ and $C$, if you like.






        share|cite|improve this answer












        Given unital $R$-algebras $Aleftarrow Bto C$, the pushout $A star_B C$ is generated as an $R$-algebra by generators of $A$ and of $C$, modulo the union of the relations in $A$ and in $C$, as well as further relations identifying the two resulting images of each element of $B$. This immediately gives the canonical maps from $A$ and $C$. You can describe this construction as a quotient of the free $R$-module on words with letters from $A$ and $C$, if you like.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 18 at 8:18









        Kevin Carlson

        32.1k23270




        32.1k23270






























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