Universal Property Of $text{Bilin}(A,A',-):text{Ab}totext{Set}$
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In class, we had to state the universal property of $text{Bilin}(A,A',-):text{Ab}totext{Set}$, the functor taking an abelian group $B$ to the set of bilinear maps $f:U(A)times U(A')to U(B)$ where $U:text{Ab}totext{Set}$ is the forgetful functor. This was the solution given (under the assumption that some representing object $Aotimes A'$ of $text{Bilin}(A,A',-)$ exists):
Let $alpha:H^{Aotimes A'}congtext{Bilin}(A,A',-)$ be the isomorphism relating the representing object to the functor. Define $tau:=alpha(Aotimes A')(text{id}_{Aotimes A'})$, then for all $Xintext{Ab}_0$ and $gintext{Bilin}(A,A',X)$ there exists a unique $bar{g}intext{Ab}(Aotimes A',X)$ such that $g=(text{Bilin}(A,A',-)(bar{g}))(tau)$.
My question: how exactly is this a universal property? nLab states that a universal property boils down to the object (or something closely associated) is initial in an appropriate auxiliary category.
What I could imagine is that this means we look at the category of dependent pairs $(X,g)insum_{text{Ab}_0}text{Bilin}(A,A',X)$ with morphisms $bar{f}:(X,g)to (Y,fcirc g)$ for any $f:Xto Y$ (is there some fancy description of this construct, maybe in terms of comma categories?) and then $(Aotimes A',tau)$ is initial.
But what interests me now, and maybe this is completely obvious, but is $text{Bilin}(A,A',-)$ also initial in it's own isomorphism class in $[text{Ab},text{Set}]$?
I'm sorry if this all is just confused gibberish. Sometimes category theory blurs my thinking way too much.
category-theory
asked Nov 16 at 15:01
fweth
1,116711
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In class, we had to state the universal property of $text{Bilin}(A,A',-):text{Ab}totext{Set}$, the functor taking an abelian group $B$ to the set of bilinear maps $f:U(A)times U(A')to U(B)$ where $U:text{Ab}totext{Set}$ is the forgetful functor. This was the solution given (under the assumption that some representing object $Aotimes A'$ of $text{Bilin}(A,A',-)$ exists):
Let $alpha:H^{Aotimes A'}congtext{Bilin}(A,A',-)$ be the isomorphism relating the representing object to the functor. Define $tau:=alpha(Aotimes A')(text{id}_{Aotimes A'})$, then for all $Xintext{Ab}_0$ and $gintext{Bilin}(A,A',X)$ there exists a unique $bar{g}intext{Ab}(Aotimes A',X)$ such that $g=(text{Bilin}(A,A',-)(bar{g}))(tau)$.
My question: how exactly is this a universal property? nLab states that a universal property boils down to the object (or something closely associated) is initial in an appropriate auxiliary category.
What I could imagine is that this means we look at the category of dependent pairs $(X,g)insum_{text{Ab}_0}text{Bilin}(A,A',X)$ with morphisms $bar{f}:(X,g)to (Y,fcirc g)$ for any $f:Xto Y$ (is there some fancy description of this construct, maybe in terms of comma categories?) and then $(Aotimes A',tau)$ is initial.
But what interests me now, and maybe this is completely obvious, but is $text{Bilin}(A,A',-)$ also initial in it's own isomorphism class in $[text{Ab},text{Set}]$?
I'm sorry if this all is just confused gibberish. Sometimes category theory blurs my thinking way too much.
category-theory
asked Nov 16 at 15:01
fweth
1,116711
2
This is the universal property of the tensor product, not of $mathrm{Bilin}$. While the description of $Aotimes A'$ is correct, and you're proving that it's the initial object of the category of elements of $mathrm{Bilin}$, your question about initiality of $mathrm{Bilin}$ itself doesn't make sense: if one element of an isomorphism class is initial, so are all the others, because there must be a unique map between all of them. I don't know of a natural universal problem solved by $mathrm{Bilin}$.
– Kevin Carlson
Nov 16 at 15:26
@KevinCarlson thanks for your answer! I see your point in 'if one element of an isomorphism class is initial, so are all the others', but how can I prove that all the other elements are initial (if they are)? Also, I feel like the definition of universal property is a bit vague, e.g. the universal property of the product $Xtimes Y$ in $C$ is a terminal object in $(Deltadownarrow(X,Y))$, but it is not the object $Xtimes Yin C_0$ per se rather than some object we recognize being essentially $Xtimes Y$.
– fweth
Nov 16 at 17:15
Being initial is invariant under isomorphism. I don't understand your second comment.
– Kevin Carlson
Nov 16 at 18:50
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In class, we had to state the universal property of $text{Bilin}(A,A',-):text{Ab}totext{Set}$, the functor taking an abelian group $B$ to the set of bilinear maps $f:U(A)times U(A')to U(B)$ where $U:text{Ab}totext{Set}$ is the forgetful functor. This was the solution given (under the assumption that some representing object $Aotimes A'$ of $text{Bilin}(A,A',-)$ exists):
Let $alpha:H^{Aotimes A'}congtext{Bilin}(A,A',-)$ be the isomorphism relating the representing object to the functor. Define $tau:=alpha(Aotimes A')(text{id}_{Aotimes A'})$, then for all $Xintext{Ab}_0$ and $gintext{Bilin}(A,A',X)$ there exists a unique $bar{g}intext{Ab}(Aotimes A',X)$ such that $g=(text{Bilin}(A,A',-)(bar{g}))(tau)$.
My question: how exactly is this a universal property? nLab states that a universal property boils down to the object (or something closely associated) is initial in an appropriate auxiliary category.
What I could imagine is that this means we look at the category of dependent pairs $(X,g)insum_{text{Ab}_0}text{Bilin}(A,A',X)$ with morphisms $bar{f}:(X,g)to (Y,fcirc g)$ for any $f:Xto Y$ (is there some fancy description of this construct, maybe in terms of comma categories?) and then $(Aotimes A',tau)$ is initial.
But what interests me now, and maybe this is completely obvious, but is $text{Bilin}(A,A',-)$ also initial in it's own isomorphism class in $[text{Ab},text{Set}]$?
I'm sorry if this all is just confused gibberish. Sometimes category theory blurs my thinking way too much.
category-theory
asked Nov 16 at 15:01
fweth
1,116711
In class, we had to state the universal property of $text{Bilin}(A,A',-):text{Ab}totext{Set}$, the functor taking an abelian group $B$ to the set of bilinear maps $f:U(A)times U(A')to U(B)$ where $U:text{Ab}totext{Set}$ is the forgetful functor. This was the solution given (under the assumption that some representing object $Aotimes A'$ of $text{Bilin}(A,A',-)$ exists):
Let $alpha:H^{Aotimes A'}congtext{Bilin}(A,A',-)$ be the isomorphism relating the representing object to the functor. Define $tau:=alpha(Aotimes A')(text{id}_{Aotimes A'})$, then for all $Xintext{Ab}_0$ and $gintext{Bilin}(A,A',X)$ there exists a unique $bar{g}intext{Ab}(Aotimes A',X)$ such that $g=(text{Bilin}(A,A',-)(bar{g}))(tau)$.
My question: how exactly is this a universal property? nLab states that a universal property boils down to the object (or something closely associated) is initial in an appropriate auxiliary category.
What I could imagine is that this means we look at the category of dependent pairs $(X,g)insum_{text{Ab}_0}text{Bilin}(A,A',X)$ with morphisms $bar{f}:(X,g)to (Y,fcirc g)$ for any $f:Xto Y$ (is there some fancy description of this construct, maybe in terms of comma categories?) and then $(Aotimes A',tau)$ is initial.
But what interests me now, and maybe this is completely obvious, but is $text{Bilin}(A,A',-)$ also initial in it's own isomorphism class in $[text{Ab},text{Set}]$?
I'm sorry if this all is just confused gibberish. Sometimes category theory blurs my thinking way too much.
category-theory
category-theory
asked Nov 16 at 15:01
fweth
1,116711
asked Nov 16 at 15:01
fweth
1,116711
asked Nov 16 at 15:01
fweth
1,116711
asked Nov 16 at 15:01
fweth
1,116711
asked Nov 16 at 15:01
fweth
1,116711
1,116711
2
This is the universal property of the tensor product, not of $mathrm{Bilin}$. While the description of $Aotimes A'$ is correct, and you're proving that it's the initial object of the category of elements of $mathrm{Bilin}$, your question about initiality of $mathrm{Bilin}$ itself doesn't make sense: if one element of an isomorphism class is initial, so are all the others, because there must be a unique map between all of them. I don't know of a natural universal problem solved by $mathrm{Bilin}$.
– Kevin Carlson
Nov 16 at 15:26
@KevinCarlson thanks for your answer! I see your point in 'if one element of an isomorphism class is initial, so are all the others', but how can I prove that all the other elements are initial (if they are)? Also, I feel like the definition of universal property is a bit vague, e.g. the universal property of the product $Xtimes Y$ in $C$ is a terminal object in $(Deltadownarrow(X,Y))$, but it is not the object $Xtimes Yin C_0$ per se rather than some object we recognize being essentially $Xtimes Y$.
– fweth
Nov 16 at 17:15
Being initial is invariant under isomorphism. I don't understand your second comment.
– Kevin Carlson
Nov 16 at 18:50
add a comment |
2
This is the universal property of the tensor product, not of $mathrm{Bilin}$. While the description of $Aotimes A'$ is correct, and you're proving that it's the initial object of the category of elements of $mathrm{Bilin}$, your question about initiality of $mathrm{Bilin}$ itself doesn't make sense: if one element of an isomorphism class is initial, so are all the others, because there must be a unique map between all of them. I don't know of a natural universal problem solved by $mathrm{Bilin}$.
– Kevin Carlson
Nov 16 at 15:26
@KevinCarlson thanks for your answer! I see your point in 'if one element of an isomorphism class is initial, so are all the others', but how can I prove that all the other elements are initial (if they are)? Also, I feel like the definition of universal property is a bit vague, e.g. the universal property of the product $Xtimes Y$ in $C$ is a terminal object in $(Deltadownarrow(X,Y))$, but it is not the object $Xtimes Yin C_0$ per se rather than some object we recognize being essentially $Xtimes Y$.
– fweth
Nov 16 at 17:15
Being initial is invariant under isomorphism. I don't understand your second comment.
– Kevin Carlson
Nov 16 at 18:50
2
2
This is the universal property of the tensor product, not of $mathrm{Bilin}$. While the description of $Aotimes A'$ is correct, and you're proving that it's the initial object of the category of elements of $mathrm{Bilin}$, your question about initiality of $mathrm{Bilin}$ itself doesn't make sense: if one element of an isomorphism class is initial, so are all the others, because there must be a unique map between all of them. I don't know of a natural universal problem solved by $mathrm{Bilin}$.
– Kevin Carlson
Nov 16 at 15:26
This is the universal property of the tensor product, not of $mathrm{Bilin}$. While the description of $Aotimes A'$ is correct, and you're proving that it's the initial object of the category of elements of $mathrm{Bilin}$, your question about initiality of $mathrm{Bilin}$ itself doesn't make sense: if one element of an isomorphism class is initial, so are all the others, because there must be a unique map between all of them. I don't know of a natural universal problem solved by $mathrm{Bilin}$.
– Kevin Carlson
Nov 16 at 15:26
@KevinCarlson thanks for your answer! I see your point in 'if one element of an isomorphism class is initial, so are all the others', but how can I prove that all the other elements are initial (if they are)? Also, I feel like the definition of universal property is a bit vague, e.g. the universal property of the product $Xtimes Y$ in $C$ is a terminal object in $(Deltadownarrow(X,Y))$, but it is not the object $Xtimes Yin C_0$ per se rather than some object we recognize being essentially $Xtimes Y$.
– fweth
Nov 16 at 17:15
@KevinCarlson thanks for your answer! I see your point in 'if one element of an isomorphism class is initial, so are all the others', but how can I prove that all the other elements are initial (if they are)? Also, I feel like the definition of universal property is a bit vague, e.g. the universal property of the product $Xtimes Y$ in $C$ is a terminal object in $(Deltadownarrow(X,Y))$, but it is not the object $Xtimes Yin C_0$ per se rather than some object we recognize being essentially $Xtimes Y$.
– fweth
Nov 16 at 17:15
Being initial is invariant under isomorphism. I don't understand your second comment.
– Kevin Carlson
Nov 16 at 18:50
Being initial is invariant under isomorphism. I don't understand your second comment.
– Kevin Carlson
Nov 16 at 18:50
add a comment |
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This is the universal property of the tensor product, not of $mathrm{Bilin}$. While the description of $Aotimes A'$ is correct, and you're proving that it's the initial object of the category of elements of $mathrm{Bilin}$, your question about initiality of $mathrm{Bilin}$ itself doesn't make sense: if one element of an isomorphism class is initial, so are all the others, because there must be a unique map between all of them. I don't know of a natural universal problem solved by $mathrm{Bilin}$.
– Kevin Carlson
Nov 16 at 15:26
@KevinCarlson thanks for your answer! I see your point in 'if one element of an isomorphism class is initial, so are all the others', but how can I prove that all the other elements are initial (if they are)? Also, I feel like the definition of universal property is a bit vague, e.g. the universal property of the product $Xtimes Y$ in $C$ is a terminal object in $(Deltadownarrow(X,Y))$, but it is not the object $Xtimes Yin C_0$ per se rather than some object we recognize being essentially $Xtimes Y$.
– fweth
Nov 16 at 17:15
Being initial is invariant under isomorphism. I don't understand your second comment.
– Kevin Carlson
Nov 16 at 18:50