Coends and adjunctions
I was reading Fosco Loregian's paper This is the co/end, my only co/friend, and here's something that I don't understand in an exercise.
The exercise is to prove that given $F: Cto D, U: Dto C$ two functors, then $Fdashv U$ if and only if for all $G: D^{op}times Cto E$ such that $int^c G(Fc, c)$ or $int^d G(d, Ud)$ makes sense, then both do and they are isomorphic, naturally in $G$.
Assuming $Fdashv U$ one can easily prove this; it's the other direction that's bugging me for the following reason. Take $G= (hom_C(-, U(-))circ tau)^{op} : D^{op}times Cto mathbf{Set}^{op}$, where $tau : Dtimes C^{op}to C^{op}times D$ is the obvious functor. $mathbf{Set}^{op}$ being cocomplete, these coends always make sense, and we have
$$int^c G(Fc,c) = int^c hom_C^{op}(c, UFc) = int_c hom_C(c,UFc) cong mathrm{Nat}(id_C, UF)$$
and also
$$int^d G(c,Ud) = int_d hom_C(Ud, Ud) cong mathrm{Nat}(U,U)$$
so if both coends are indeed isomorphic, $mathrm{Nat}(U,U) cong mathrm{Nat}(id_C,UF)$. But this is odd because in an adjunction, what we actually get is something like $mathrm{Nat}(id_C, UF) cong mathrm{Nat}(F,F)$, not $mathrm{Nat}(U,U)$, which will rather be isomorphic to $mathrm{Nat}(FU, id_D)$.
Now I don't know if I made a mistake in my calculation, or simply I just found out something I didn't know about adjunctions. Which is it ? (I think I made a mistake at some point, probably when going from $int^d$ to $int_d$ but I don't see how: if I'm not mistaken, $int^c T(c,c) = int_c T^{op}(c,c)$ where, if $T: A^{op}times Ato C$, $T^{op} : (A^{op})^{op}times A^{op} = Atimes A^{op}to C^{op}$)
category-theory limits-colimits adjoint-functors
edited Nov 27 at 12:37
Martin Sleziak
44.7k7115270
asked Nov 2 at 12:31
Max
12.8k11040
add a comment |
I was reading Fosco Loregian's paper This is the co/end, my only co/friend, and here's something that I don't understand in an exercise.
The exercise is to prove that given $F: Cto D, U: Dto C$ two functors, then $Fdashv U$ if and only if for all $G: D^{op}times Cto E$ such that $int^c G(Fc, c)$ or $int^d G(d, Ud)$ makes sense, then both do and they are isomorphic, naturally in $G$.
Assuming $Fdashv U$ one can easily prove this; it's the other direction that's bugging me for the following reason. Take $G= (hom_C(-, U(-))circ tau)^{op} : D^{op}times Cto mathbf{Set}^{op}$, where $tau : Dtimes C^{op}to C^{op}times D$ is the obvious functor. $mathbf{Set}^{op}$ being cocomplete, these coends always make sense, and we have
$$int^c G(Fc,c) = int^c hom_C^{op}(c, UFc) = int_c hom_C(c,UFc) cong mathrm{Nat}(id_C, UF)$$
and also
$$int^d G(c,Ud) = int_d hom_C(Ud, Ud) cong mathrm{Nat}(U,U)$$
so if both coends are indeed isomorphic, $mathrm{Nat}(U,U) cong mathrm{Nat}(id_C,UF)$. But this is odd because in an adjunction, what we actually get is something like $mathrm{Nat}(id_C, UF) cong mathrm{Nat}(F,F)$, not $mathrm{Nat}(U,U)$, which will rather be isomorphic to $mathrm{Nat}(FU, id_D)$.
Now I don't know if I made a mistake in my calculation, or simply I just found out something I didn't know about adjunctions. Which is it ? (I think I made a mistake at some point, probably when going from $int^d$ to $int_d$ but I don't see how: if I'm not mistaken, $int^c T(c,c) = int_c T^{op}(c,c)$ where, if $T: A^{op}times Ato C$, $T^{op} : (A^{op})^{op}times A^{op} = Atimes A^{op}to C^{op}$)
category-theory limits-colimits adjoint-functors
edited Nov 27 at 12:37
Martin Sleziak
44.7k7115270
asked Nov 2 at 12:31
Max
12.8k11040
1
All these are isomorphic, see math.stackexchange.com/questions/1370444/adjoint-squares
– Arnaud D.
Nov 2 at 13:54
@ArnaudD. thank you for the very useful link, so my formula gives another proof if I made no mistake, right? (of course this proof is less illuminating than the one in the link you provided) (I started "fearing" that they were indeed isomorphic when looking at examples, free-forgetful adjunctions from algebraic categories to $mathbf{Set}$ are particularly illuminating for instance)
– Max
Nov 2 at 14:19
Yes, I think the results in the link can be recovered as special case of the one you mention here.
– Arnaud D.
Nov 3 at 13:48
@ArnaudD. : oh right I hadn't noticed the tag limits and colimits, I think you're right - how do I delete it ?
– Max
Nov 3 at 15:34
add a comment |
I was reading Fosco Loregian's paper This is the co/end, my only co/friend, and here's something that I don't understand in an exercise.
The exercise is to prove that given $F: Cto D, U: Dto C$ two functors, then $Fdashv U$ if and only if for all $G: D^{op}times Cto E$ such that $int^c G(Fc, c)$ or $int^d G(d, Ud)$ makes sense, then both do and they are isomorphic, naturally in $G$.
Assuming $Fdashv U$ one can easily prove this; it's the other direction that's bugging me for the following reason. Take $G= (hom_C(-, U(-))circ tau)^{op} : D^{op}times Cto mathbf{Set}^{op}$, where $tau : Dtimes C^{op}to C^{op}times D$ is the obvious functor. $mathbf{Set}^{op}$ being cocomplete, these coends always make sense, and we have
$$int^c G(Fc,c) = int^c hom_C^{op}(c, UFc) = int_c hom_C(c,UFc) cong mathrm{Nat}(id_C, UF)$$
and also
$$int^d G(c,Ud) = int_d hom_C(Ud, Ud) cong mathrm{Nat}(U,U)$$
so if both coends are indeed isomorphic, $mathrm{Nat}(U,U) cong mathrm{Nat}(id_C,UF)$. But this is odd because in an adjunction, what we actually get is something like $mathrm{Nat}(id_C, UF) cong mathrm{Nat}(F,F)$, not $mathrm{Nat}(U,U)$, which will rather be isomorphic to $mathrm{Nat}(FU, id_D)$.
Now I don't know if I made a mistake in my calculation, or simply I just found out something I didn't know about adjunctions. Which is it ? (I think I made a mistake at some point, probably when going from $int^d$ to $int_d$ but I don't see how: if I'm not mistaken, $int^c T(c,c) = int_c T^{op}(c,c)$ where, if $T: A^{op}times Ato C$, $T^{op} : (A^{op})^{op}times A^{op} = Atimes A^{op}to C^{op}$)
category-theory limits-colimits adjoint-functors
edited Nov 27 at 12:37
Martin Sleziak
44.7k7115270
asked Nov 2 at 12:31
Max
12.8k11040
I was reading Fosco Loregian's paper This is the co/end, my only co/friend, and here's something that I don't understand in an exercise.
The exercise is to prove that given $F: Cto D, U: Dto C$ two functors, then $Fdashv U$ if and only if for all $G: D^{op}times Cto E$ such that $int^c G(Fc, c)$ or $int^d G(d, Ud)$ makes sense, then both do and they are isomorphic, naturally in $G$.
Assuming $Fdashv U$ one can easily prove this; it's the other direction that's bugging me for the following reason. Take $G= (hom_C(-, U(-))circ tau)^{op} : D^{op}times Cto mathbf{Set}^{op}$, where $tau : Dtimes C^{op}to C^{op}times D$ is the obvious functor. $mathbf{Set}^{op}$ being cocomplete, these coends always make sense, and we have
$$int^c G(Fc,c) = int^c hom_C^{op}(c, UFc) = int_c hom_C(c,UFc) cong mathrm{Nat}(id_C, UF)$$
and also
$$int^d G(c,Ud) = int_d hom_C(Ud, Ud) cong mathrm{Nat}(U,U)$$
so if both coends are indeed isomorphic, $mathrm{Nat}(U,U) cong mathrm{Nat}(id_C,UF)$. But this is odd because in an adjunction, what we actually get is something like $mathrm{Nat}(id_C, UF) cong mathrm{Nat}(F,F)$, not $mathrm{Nat}(U,U)$, which will rather be isomorphic to $mathrm{Nat}(FU, id_D)$.
Now I don't know if I made a mistake in my calculation, or simply I just found out something I didn't know about adjunctions. Which is it ? (I think I made a mistake at some point, probably when going from $int^d$ to $int_d$ but I don't see how: if I'm not mistaken, $int^c T(c,c) = int_c T^{op}(c,c)$ where, if $T: A^{op}times Ato C$, $T^{op} : (A^{op})^{op}times A^{op} = Atimes A^{op}to C^{op}$)
category-theory limits-colimits adjoint-functors
category-theory limits-colimits adjoint-functors
edited Nov 27 at 12:37
Martin Sleziak
44.7k7115270
asked Nov 2 at 12:31
Max
12.8k11040
edited Nov 27 at 12:37
Martin Sleziak
44.7k7115270
asked Nov 2 at 12:31
Max
12.8k11040
edited Nov 27 at 12:37
Martin Sleziak
44.7k7115270
edited Nov 27 at 12:37
Martin Sleziak
44.7k7115270
edited Nov 27 at 12:37
Martin Sleziak
44.7k7115270
44.7k7115270
asked Nov 2 at 12:31
Max
12.8k11040
asked Nov 2 at 12:31
Max
12.8k11040
asked Nov 2 at 12:31
Max
12.8k11040
12.8k11040
1
All these are isomorphic, see math.stackexchange.com/questions/1370444/adjoint-squares
– Arnaud D.
Nov 2 at 13:54
@ArnaudD. thank you for the very useful link, so my formula gives another proof if I made no mistake, right? (of course this proof is less illuminating than the one in the link you provided) (I started "fearing" that they were indeed isomorphic when looking at examples, free-forgetful adjunctions from algebraic categories to $mathbf{Set}$ are particularly illuminating for instance)
– Max
Nov 2 at 14:19
Yes, I think the results in the link can be recovered as special case of the one you mention here.
– Arnaud D.
Nov 3 at 13:48
@ArnaudD. : oh right I hadn't noticed the tag limits and colimits, I think you're right - how do I delete it ?
– Max
Nov 3 at 15:34
add a comment |
1
All these are isomorphic, see math.stackexchange.com/questions/1370444/adjoint-squares
– Arnaud D.
Nov 2 at 13:54
@ArnaudD. thank you for the very useful link, so my formula gives another proof if I made no mistake, right? (of course this proof is less illuminating than the one in the link you provided) (I started "fearing" that they were indeed isomorphic when looking at examples, free-forgetful adjunctions from algebraic categories to $mathbf{Set}$ are particularly illuminating for instance)
– Max
Nov 2 at 14:19
Yes, I think the results in the link can be recovered as special case of the one you mention here.
– Arnaud D.
Nov 3 at 13:48
@ArnaudD. : oh right I hadn't noticed the tag limits and colimits, I think you're right - how do I delete it ?
– Max
Nov 3 at 15:34
1
1
All these are isomorphic, see math.stackexchange.com/questions/1370444/adjoint-squares
– Arnaud D.
Nov 2 at 13:54
All these are isomorphic, see math.stackexchange.com/questions/1370444/adjoint-squares
– Arnaud D.
Nov 2 at 13:54
@ArnaudD. thank you for the very useful link, so my formula gives another proof if I made no mistake, right? (of course this proof is less illuminating than the one in the link you provided) (I started "fearing" that they were indeed isomorphic when looking at examples, free-forgetful adjunctions from algebraic categories to $mathbf{Set}$ are particularly illuminating for instance)
– Max
Nov 2 at 14:19
@ArnaudD. thank you for the very useful link, so my formula gives another proof if I made no mistake, right? (of course this proof is less illuminating than the one in the link you provided) (I started "fearing" that they were indeed isomorphic when looking at examples, free-forgetful adjunctions from algebraic categories to $mathbf{Set}$ are particularly illuminating for instance)
– Max
Nov 2 at 14:19
Yes, I think the results in the link can be recovered as special case of the one you mention here.
– Arnaud D.
Nov 3 at 13:48
Yes, I think the results in the link can be recovered as special case of the one you mention here.
– Arnaud D.
Nov 3 at 13:48
@ArnaudD. : oh right I hadn't noticed the tag limits and colimits, I think you're right - how do I delete it ?
– Max
Nov 3 at 15:34
@ArnaudD. : oh right I hadn't noticed the tag limits and colimits, I think you're right - how do I delete it ?
– Max
Nov 3 at 15:34
add a comment |
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All these are isomorphic, see math.stackexchange.com/questions/1370444/adjoint-squares
– Arnaud D.
Nov 2 at 13:54
@ArnaudD. thank you for the very useful link, so my formula gives another proof if I made no mistake, right? (of course this proof is less illuminating than the one in the link you provided) (I started "fearing" that they were indeed isomorphic when looking at examples, free-forgetful adjunctions from algebraic categories to $mathbf{Set}$ are particularly illuminating for instance)
– Max
Nov 2 at 14:19
Yes, I think the results in the link can be recovered as special case of the one you mention here.
– Arnaud D.
Nov 3 at 13:48
@ArnaudD. : oh right I hadn't noticed the tag limits and colimits, I think you're right - how do I delete it ?
– Max
Nov 3 at 15:34