Mixing
Left and Right adjoint of forgetful functor
I am studying category theory and I came across two questions that I could not answer about the adjoint of two forgetful functors.
Let $mathbf{TAb}$ be the category of abelian groups such that all elements have finite order and let $mathbf{Grp}$ be the category of groups. Why does the forgetful functor $U_1 : mathbf{TAb}rightarrow mathbf{Set}$ does not have left adjoint? And why the forgetful functor $U_2: mathbf{Grp} rightarrow mathbf{Set}$ does not have right adjoint?
The autor also asks about the right adjoint of the forgetful functors $mathbf{k}text{-}mathbf{Mod} rightarrow mathbf{Set}$ and $mathbf{Ring}rightarrow mathbf{Set}$, but I think that they don't have right adjoint for the same reason of $U_2$.
category-theory
asked Nov 20 '18 at 18:51
H R
1608
add a comment |
I am studying category theory and I came across two questions that I could not answer about the adjoint of two forgetful functors.
Let $mathbf{TAb}$ be the category of abelian groups such that all elements have finite order and let $mathbf{Grp}$ be the category of groups. Why does the forgetful functor $U_1 : mathbf{TAb}rightarrow mathbf{Set}$ does not have left adjoint? And why the forgetful functor $U_2: mathbf{Grp} rightarrow mathbf{Set}$ does not have right adjoint?
The autor also asks about the right adjoint of the forgetful functors $mathbf{k}text{-}mathbf{Mod} rightarrow mathbf{Set}$ and $mathbf{Ring}rightarrow mathbf{Set}$, but I think that they don't have right adjoint for the same reason of $U_2$.
category-theory
asked Nov 20 '18 at 18:51
H R
1608
add a comment |
I am studying category theory and I came across two questions that I could not answer about the adjoint of two forgetful functors.
Let $mathbf{TAb}$ be the category of abelian groups such that all elements have finite order and let $mathbf{Grp}$ be the category of groups. Why does the forgetful functor $U_1 : mathbf{TAb}rightarrow mathbf{Set}$ does not have left adjoint? And why the forgetful functor $U_2: mathbf{Grp} rightarrow mathbf{Set}$ does not have right adjoint?
The autor also asks about the right adjoint of the forgetful functors $mathbf{k}text{-}mathbf{Mod} rightarrow mathbf{Set}$ and $mathbf{Ring}rightarrow mathbf{Set}$, but I think that they don't have right adjoint for the same reason of $U_2$.
category-theory
asked Nov 20 '18 at 18:51
H R
1608
I am studying category theory and I came across two questions that I could not answer about the adjoint of two forgetful functors.
Let $mathbf{TAb}$ be the category of abelian groups such that all elements have finite order and let $mathbf{Grp}$ be the category of groups. Why does the forgetful functor $U_1 : mathbf{TAb}rightarrow mathbf{Set}$ does not have left adjoint? And why the forgetful functor $U_2: mathbf{Grp} rightarrow mathbf{Set}$ does not have right adjoint?
The autor also asks about the right adjoint of the forgetful functors $mathbf{k}text{-}mathbf{Mod} rightarrow mathbf{Set}$ and $mathbf{Ring}rightarrow mathbf{Set}$, but I think that they don't have right adjoint for the same reason of $U_2$.
category-theory
category-theory
asked Nov 20 '18 at 18:51
H R
1608
asked Nov 20 '18 at 18:51
H R
1608
asked Nov 20 '18 at 18:51
H R
1608
asked Nov 20 '18 at 18:51
H R
1608
asked Nov 20 '18 at 18:51
H R
1608
1608
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
A very interesting property of adjoints is that they preserve (co)limits: right adjoints preserve limits, left adjoints preserve colimits.
Take for instance a family of torsion abelian groups $G_i, iin I$, with $|I|geq 2$ then their coproduct in $mathbf{TAb}$ is $displaystylebigoplus_{iin I}G_i$, while the coproduct of $U_1(G_i), iin I$ is $displaystylecoprod_{iin I}U_1(G_i)$ : $U_1$ doesn't preserve colimits, therefore it can't be a left adjoint, i.e. it can't have a right adjoint.
You can do a very similar proof for $U_2$ (the coproduct looks a bit different in $mathbf{Grp}$ but it's still not preserved by $U_2$)
You can wonder whether this is a good criterion and Freyd's adjoint theorem essentially says that it is, in that for nice categories and nice functors, preserving (co)limits is enough to be a (left) right adjoint
answered Nov 20 '18 at 20:24
Max
12.9k11040
So to prove that $U_1$ doesn't have a left adjoint, do I just need to prove that it does not preserve limits?
– H R
Nov 20 '18 at 22:28
Yes for instance you can do that
– Max
Nov 20 '18 at 22:40
add a comment |
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A very interesting property of adjoints is that they preserve (co)limits: right adjoints preserve limits, left adjoints preserve colimits.
Take for instance a family of torsion abelian groups $G_i, iin I$, with $|I|geq 2$ then their coproduct in $mathbf{TAb}$ is $displaystylebigoplus_{iin I}G_i$, while the coproduct of $U_1(G_i), iin I$ is $displaystylecoprod_{iin I}U_1(G_i)$ : $U_1$ doesn't preserve colimits, therefore it can't be a left adjoint, i.e. it can't have a right adjoint.
You can do a very similar proof for $U_2$ (the coproduct looks a bit different in $mathbf{Grp}$ but it's still not preserved by $U_2$)
You can wonder whether this is a good criterion and Freyd's adjoint theorem essentially says that it is, in that for nice categories and nice functors, preserving (co)limits is enough to be a (left) right adjoint
answered Nov 20 '18 at 20:24
Max
12.9k11040
So to prove that $U_1$ doesn't have a left adjoint, do I just need to prove that it does not preserve limits?
– H R
Nov 20 '18 at 22:28
Yes for instance you can do that
– Max
Nov 20 '18 at 22:40
add a comment |
A very interesting property of adjoints is that they preserve (co)limits: right adjoints preserve limits, left adjoints preserve colimits.
Take for instance a family of torsion abelian groups $G_i, iin I$, with $|I|geq 2$ then their coproduct in $mathbf{TAb}$ is $displaystylebigoplus_{iin I}G_i$, while the coproduct of $U_1(G_i), iin I$ is $displaystylecoprod_{iin I}U_1(G_i)$ : $U_1$ doesn't preserve colimits, therefore it can't be a left adjoint, i.e. it can't have a right adjoint.
You can do a very similar proof for $U_2$ (the coproduct looks a bit different in $mathbf{Grp}$ but it's still not preserved by $U_2$)
You can wonder whether this is a good criterion and Freyd's adjoint theorem essentially says that it is, in that for nice categories and nice functors, preserving (co)limits is enough to be a (left) right adjoint
answered Nov 20 '18 at 20:24
Max
12.9k11040
So to prove that $U_1$ doesn't have a left adjoint, do I just need to prove that it does not preserve limits?
– H R
Nov 20 '18 at 22:28
Yes for instance you can do that
– Max
Nov 20 '18 at 22:40
add a comment |
A very interesting property of adjoints is that they preserve (co)limits: right adjoints preserve limits, left adjoints preserve colimits.
Take for instance a family of torsion abelian groups $G_i, iin I$, with $|I|geq 2$ then their coproduct in $mathbf{TAb}$ is $displaystylebigoplus_{iin I}G_i$, while the coproduct of $U_1(G_i), iin I$ is $displaystylecoprod_{iin I}U_1(G_i)$ : $U_1$ doesn't preserve colimits, therefore it can't be a left adjoint, i.e. it can't have a right adjoint.
You can do a very similar proof for $U_2$ (the coproduct looks a bit different in $mathbf{Grp}$ but it's still not preserved by $U_2$)
You can wonder whether this is a good criterion and Freyd's adjoint theorem essentially says that it is, in that for nice categories and nice functors, preserving (co)limits is enough to be a (left) right adjoint
answered Nov 20 '18 at 20:24
Max
12.9k11040
A very interesting property of adjoints is that they preserve (co)limits: right adjoints preserve limits, left adjoints preserve colimits.
Take for instance a family of torsion abelian groups $G_i, iin I$, with $|I|geq 2$ then their coproduct in $mathbf{TAb}$ is $displaystylebigoplus_{iin I}G_i$, while the coproduct of $U_1(G_i), iin I$ is $displaystylecoprod_{iin I}U_1(G_i)$ : $U_1$ doesn't preserve colimits, therefore it can't be a left adjoint, i.e. it can't have a right adjoint.
You can do a very similar proof for $U_2$ (the coproduct looks a bit different in $mathbf{Grp}$ but it's still not preserved by $U_2$)
You can wonder whether this is a good criterion and Freyd's adjoint theorem essentially says that it is, in that for nice categories and nice functors, preserving (co)limits is enough to be a (left) right adjoint
answered Nov 20 '18 at 20:24
Max
12.9k11040
answered Nov 20 '18 at 20:24
Max
12.9k11040
answered Nov 20 '18 at 20:24
Max
12.9k11040
answered Nov 20 '18 at 20:24
Max
12.9k11040
12.9k11040
So to prove that $U_1$ doesn't have a left adjoint, do I just need to prove that it does not preserve limits?
– H R
Nov 20 '18 at 22:28
Yes for instance you can do that
– Max
Nov 20 '18 at 22:40
add a comment |
So to prove that $U_1$ doesn't have a left adjoint, do I just need to prove that it does not preserve limits?
– H R
Nov 20 '18 at 22:28
Yes for instance you can do that
– Max
Nov 20 '18 at 22:40
So to prove that $U_1$ doesn't have a left adjoint, do I just need to prove that it does not preserve limits?
– H R
Nov 20 '18 at 22:28
So to prove that $U_1$ doesn't have a left adjoint, do I just need to prove that it does not preserve limits?
– H R
Nov 20 '18 at 22:28
Yes for instance you can do that
– Max
Nov 20 '18 at 22:40
Yes for instance you can do that
– Max
Nov 20 '18 at 22:40
add a comment |
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