Is my understanding of the diagonal functor correct, and if so, what's the point?












1














Here's a definition of the diagonal functor from my lecture notes:




Let us introduce the diagonal functor $Delta: C rightarrow operatorname{Fun}(I, C)$ given by sending $x in C$ to the constant functor $i mapsto x$ (and morphisms sent to identities).




So define the constant functor $F_x: I rightarrow C$ that maps every $i in operatorname{Ob} I$ to $x in operatorname{Ob} C$ and every morphism $i rightarrow j in I$ to $id_x: x rightarrow x$.



Then consider morphisms $f: x rightarrow y$ in $C$ and $phi: i rightarrow j$ in $I$. According to the above definition, $Delta$ maps $x mapsto F_x, y mapsto F_y$ and it maps $f: xrightarrow y$ to some $Delta_f: F_x(i) rightarrow F_y(j)$. So I guess when the definition above says "morphisms sent to identities" what they mean is that $Delta_f(i) = f$ for all $i in I$.



Then the square with upper row $x = F_x(i) xrightarrow{F_x(phi)=id_x} F_x(j) = x$, and lower row $y = F_y(i) xrightarrow{F_y(phi)=id_y} F_y(j) = y$, and vertical maps $Delta_f(i) = f$ and $Delta_f(j) = f$ commutes trivially, ie. this does seem to be a well-defined functor (modulo a few additional checks, on compositions etc).



Is the above interpretation correct? If so, what motivates this? The square I just constructed is not too illuminating...






category-theory







share|cite|improve this question
























  • "Morphisms sent to identities" is referring to the functor that is produced by the diagonal functor. That is, it is saying $Delta(x)(phi)=id_x$ it is not referring to $Delta$'s action on arrows.
    – Derek Elkins
    Jan 4 at 18:25










  • Conceptually, many important constructs are revealed in category theory to be determined (usually via adjunction) from "trivial"-seeming functors.
    – Derek Elkins
    Jan 4 at 18:30










  • @DerekElkins i was very confused abt the statement in brackets, thx for clarifying
    – gen
    Jan 4 at 18:39
















1














Here's a definition of the diagonal functor from my lecture notes:




Let us introduce the diagonal functor $Delta: C rightarrow operatorname{Fun}(I, C)$ given by sending $x in C$ to the constant functor $i mapsto x$ (and morphisms sent to identities).




So define the constant functor $F_x: I rightarrow C$ that maps every $i in operatorname{Ob} I$ to $x in operatorname{Ob} C$ and every morphism $i rightarrow j in I$ to $id_x: x rightarrow x$.



Then consider morphisms $f: x rightarrow y$ in $C$ and $phi: i rightarrow j$ in $I$. According to the above definition, $Delta$ maps $x mapsto F_x, y mapsto F_y$ and it maps $f: xrightarrow y$ to some $Delta_f: F_x(i) rightarrow F_y(j)$. So I guess when the definition above says "morphisms sent to identities" what they mean is that $Delta_f(i) = f$ for all $i in I$.



Then the square with upper row $x = F_x(i) xrightarrow{F_x(phi)=id_x} F_x(j) = x$, and lower row $y = F_y(i) xrightarrow{F_y(phi)=id_y} F_y(j) = y$, and vertical maps $Delta_f(i) = f$ and $Delta_f(j) = f$ commutes trivially, ie. this does seem to be a well-defined functor (modulo a few additional checks, on compositions etc).



Is the above interpretation correct? If so, what motivates this? The square I just constructed is not too illuminating...






category-theory







share|cite|improve this question
























  • "Morphisms sent to identities" is referring to the functor that is produced by the diagonal functor. That is, it is saying $Delta(x)(phi)=id_x$ it is not referring to $Delta$'s action on arrows.
    – Derek Elkins
    Jan 4 at 18:25










  • Conceptually, many important constructs are revealed in category theory to be determined (usually via adjunction) from "trivial"-seeming functors.
    – Derek Elkins
    Jan 4 at 18:30










  • @DerekElkins i was very confused abt the statement in brackets, thx for clarifying
    – gen
    Jan 4 at 18:39














1












1








1







Here's a definition of the diagonal functor from my lecture notes:




Let us introduce the diagonal functor $Delta: C rightarrow operatorname{Fun}(I, C)$ given by sending $x in C$ to the constant functor $i mapsto x$ (and morphisms sent to identities).




So define the constant functor $F_x: I rightarrow C$ that maps every $i in operatorname{Ob} I$ to $x in operatorname{Ob} C$ and every morphism $i rightarrow j in I$ to $id_x: x rightarrow x$.



Then consider morphisms $f: x rightarrow y$ in $C$ and $phi: i rightarrow j$ in $I$. According to the above definition, $Delta$ maps $x mapsto F_x, y mapsto F_y$ and it maps $f: xrightarrow y$ to some $Delta_f: F_x(i) rightarrow F_y(j)$. So I guess when the definition above says "morphisms sent to identities" what they mean is that $Delta_f(i) = f$ for all $i in I$.



Then the square with upper row $x = F_x(i) xrightarrow{F_x(phi)=id_x} F_x(j) = x$, and lower row $y = F_y(i) xrightarrow{F_y(phi)=id_y} F_y(j) = y$, and vertical maps $Delta_f(i) = f$ and $Delta_f(j) = f$ commutes trivially, ie. this does seem to be a well-defined functor (modulo a few additional checks, on compositions etc).



Is the above interpretation correct? If so, what motivates this? The square I just constructed is not too illuminating...






category-theory







share|cite|improve this question















Here's a definition of the diagonal functor from my lecture notes:




Let us introduce the diagonal functor $Delta: C rightarrow operatorname{Fun}(I, C)$ given by sending $x in C$ to the constant functor $i mapsto x$ (and morphisms sent to identities).




So define the constant functor $F_x: I rightarrow C$ that maps every $i in operatorname{Ob} I$ to $x in operatorname{Ob} C$ and every morphism $i rightarrow j in I$ to $id_x: x rightarrow x$.



Then consider morphisms $f: x rightarrow y$ in $C$ and $phi: i rightarrow j$ in $I$. According to the above definition, $Delta$ maps $x mapsto F_x, y mapsto F_y$ and it maps $f: xrightarrow y$ to some $Delta_f: F_x(i) rightarrow F_y(j)$. So I guess when the definition above says "morphisms sent to identities" what they mean is that $Delta_f(i) = f$ for all $i in I$.



Then the square with upper row $x = F_x(i) xrightarrow{F_x(phi)=id_x} F_x(j) = x$, and lower row $y = F_y(i) xrightarrow{F_y(phi)=id_y} F_y(j) = y$, and vertical maps $Delta_f(i) = f$ and $Delta_f(j) = f$ commutes trivially, ie. this does seem to be a well-defined functor (modulo a few additional checks, on compositions etc).



Is the above interpretation correct? If so, what motivates this? The square I just constructed is not too illuminating...






category-theory




category-theory






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edited Jan 4 at 13:27









Shaun

8,820113681




8,820113681










asked Jan 4 at 12:14









gengen

4382521




4382521












  • "Morphisms sent to identities" is referring to the functor that is produced by the diagonal functor. That is, it is saying $Delta(x)(phi)=id_x$ it is not referring to $Delta$'s action on arrows.
    – Derek Elkins
    Jan 4 at 18:25










  • Conceptually, many important constructs are revealed in category theory to be determined (usually via adjunction) from "trivial"-seeming functors.
    – Derek Elkins
    Jan 4 at 18:30










  • @DerekElkins i was very confused abt the statement in brackets, thx for clarifying
    – gen
    Jan 4 at 18:39


















  • "Morphisms sent to identities" is referring to the functor that is produced by the diagonal functor. That is, it is saying $Delta(x)(phi)=id_x$ it is not referring to $Delta$'s action on arrows.
    – Derek Elkins
    Jan 4 at 18:25










  • Conceptually, many important constructs are revealed in category theory to be determined (usually via adjunction) from "trivial"-seeming functors.
    – Derek Elkins
    Jan 4 at 18:30










  • @DerekElkins i was very confused abt the statement in brackets, thx for clarifying
    – gen
    Jan 4 at 18:39
















"Morphisms sent to identities" is referring to the functor that is produced by the diagonal functor. That is, it is saying $Delta(x)(phi)=id_x$ it is not referring to $Delta$'s action on arrows.
– Derek Elkins
Jan 4 at 18:25




"Morphisms sent to identities" is referring to the functor that is produced by the diagonal functor. That is, it is saying $Delta(x)(phi)=id_x$ it is not referring to $Delta$'s action on arrows.
– Derek Elkins
Jan 4 at 18:25












Conceptually, many important constructs are revealed in category theory to be determined (usually via adjunction) from "trivial"-seeming functors.
– Derek Elkins
Jan 4 at 18:30




Conceptually, many important constructs are revealed in category theory to be determined (usually via adjunction) from "trivial"-seeming functors.
– Derek Elkins
Jan 4 at 18:30












@DerekElkins i was very confused abt the statement in brackets, thx for clarifying
– gen
Jan 4 at 18:39




@DerekElkins i was very confused abt the statement in brackets, thx for clarifying
– gen
Jan 4 at 18:39










1 Answer
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Your interpretation is correct.



It seems like a trivial functor, but it's actually very useful—for example, a category $mathcal{C}$ has all limits (resp. colimits) of shape $mathcal{I}$ if and only if the diagonal functor $Delta : mathcal{C} to mathrm{Fun}(mathcal{I}, mathcal{C})$ has a right (resp. left) adjoint:
$$mathrm{colim} dashv Delta dashv mathrm{lim}$$



A specific instance of this is when $mathcal{I} = mathbf{0}$ is the empty category. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathbf{1}$, the terminal category, and then the corresponding diagonal functor is just the unique functor $mathcal{C} to mathbf{1}$. This functor has a right (resp. left) adjoint if and only if $mathcal{C}$ has a terminal (resp. initial) object.



Another instance is when $mathcal{I} = 2$ is the discrete category with two objects. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathcal{C}^2$, and the diagonal functor $mathcal{C} to mathcal{C}^2$ has a right (resp. left) adjoint if and only if $mathcal{C}$ has binary products (resp. coproducts).






share|cite|improve this answer



















  • 1




    I've reversed Shaun's edit. I used unbolded $2$ on purpose for the discrete category with two objects, to distinguish it from $mathbf{2}$, which is the category with two objects and a unique morphism between them.
    – Clive Newstead
    Jan 4 at 13:49










  • I often use $1+1$ to avoid such ambiguity.
    – Derek Elkins
    Jan 4 at 18:30













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1 Answer
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2














Your interpretation is correct.



It seems like a trivial functor, but it's actually very useful—for example, a category $mathcal{C}$ has all limits (resp. colimits) of shape $mathcal{I}$ if and only if the diagonal functor $Delta : mathcal{C} to mathrm{Fun}(mathcal{I}, mathcal{C})$ has a right (resp. left) adjoint:
$$mathrm{colim} dashv Delta dashv mathrm{lim}$$



A specific instance of this is when $mathcal{I} = mathbf{0}$ is the empty category. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathbf{1}$, the terminal category, and then the corresponding diagonal functor is just the unique functor $mathcal{C} to mathbf{1}$. This functor has a right (resp. left) adjoint if and only if $mathcal{C}$ has a terminal (resp. initial) object.



Another instance is when $mathcal{I} = 2$ is the discrete category with two objects. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathcal{C}^2$, and the diagonal functor $mathcal{C} to mathcal{C}^2$ has a right (resp. left) adjoint if and only if $mathcal{C}$ has binary products (resp. coproducts).






share|cite|improve this answer



















  • 1




    I've reversed Shaun's edit. I used unbolded $2$ on purpose for the discrete category with two objects, to distinguish it from $mathbf{2}$, which is the category with two objects and a unique morphism between them.
    – Clive Newstead
    Jan 4 at 13:49










  • I often use $1+1$ to avoid such ambiguity.
    – Derek Elkins
    Jan 4 at 18:30


















2














Your interpretation is correct.



It seems like a trivial functor, but it's actually very useful—for example, a category $mathcal{C}$ has all limits (resp. colimits) of shape $mathcal{I}$ if and only if the diagonal functor $Delta : mathcal{C} to mathrm{Fun}(mathcal{I}, mathcal{C})$ has a right (resp. left) adjoint:
$$mathrm{colim} dashv Delta dashv mathrm{lim}$$



A specific instance of this is when $mathcal{I} = mathbf{0}$ is the empty category. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathbf{1}$, the terminal category, and then the corresponding diagonal functor is just the unique functor $mathcal{C} to mathbf{1}$. This functor has a right (resp. left) adjoint if and only if $mathcal{C}$ has a terminal (resp. initial) object.



Another instance is when $mathcal{I} = 2$ is the discrete category with two objects. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathcal{C}^2$, and the diagonal functor $mathcal{C} to mathcal{C}^2$ has a right (resp. left) adjoint if and only if $mathcal{C}$ has binary products (resp. coproducts).






share|cite|improve this answer



















  • 1




    I've reversed Shaun's edit. I used unbolded $2$ on purpose for the discrete category with two objects, to distinguish it from $mathbf{2}$, which is the category with two objects and a unique morphism between them.
    – Clive Newstead
    Jan 4 at 13:49










  • I often use $1+1$ to avoid such ambiguity.
    – Derek Elkins
    Jan 4 at 18:30
















2












2








2






Your interpretation is correct.



It seems like a trivial functor, but it's actually very useful—for example, a category $mathcal{C}$ has all limits (resp. colimits) of shape $mathcal{I}$ if and only if the diagonal functor $Delta : mathcal{C} to mathrm{Fun}(mathcal{I}, mathcal{C})$ has a right (resp. left) adjoint:
$$mathrm{colim} dashv Delta dashv mathrm{lim}$$



A specific instance of this is when $mathcal{I} = mathbf{0}$ is the empty category. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathbf{1}$, the terminal category, and then the corresponding diagonal functor is just the unique functor $mathcal{C} to mathbf{1}$. This functor has a right (resp. left) adjoint if and only if $mathcal{C}$ has a terminal (resp. initial) object.



Another instance is when $mathcal{I} = 2$ is the discrete category with two objects. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathcal{C}^2$, and the diagonal functor $mathcal{C} to mathcal{C}^2$ has a right (resp. left) adjoint if and only if $mathcal{C}$ has binary products (resp. coproducts).






share|cite|improve this answer














Your interpretation is correct.



It seems like a trivial functor, but it's actually very useful—for example, a category $mathcal{C}$ has all limits (resp. colimits) of shape $mathcal{I}$ if and only if the diagonal functor $Delta : mathcal{C} to mathrm{Fun}(mathcal{I}, mathcal{C})$ has a right (resp. left) adjoint:
$$mathrm{colim} dashv Delta dashv mathrm{lim}$$



A specific instance of this is when $mathcal{I} = mathbf{0}$ is the empty category. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathbf{1}$, the terminal category, and then the corresponding diagonal functor is just the unique functor $mathcal{C} to mathbf{1}$. This functor has a right (resp. left) adjoint if and only if $mathcal{C}$ has a terminal (resp. initial) object.



Another instance is when $mathcal{I} = 2$ is the discrete category with two objects. Then $mathrm{Fun}(mathcal{I}, mathcal{C}) cong mathcal{C}^2$, and the diagonal functor $mathcal{C} to mathcal{C}^2$ has a right (resp. left) adjoint if and only if $mathcal{C}$ has binary products (resp. coproducts).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 13:48

























answered Jan 4 at 13:03









Clive NewsteadClive Newstead

50.7k474133




50.7k474133








  • 1




    I've reversed Shaun's edit. I used unbolded $2$ on purpose for the discrete category with two objects, to distinguish it from $mathbf{2}$, which is the category with two objects and a unique morphism between them.
    – Clive Newstead
    Jan 4 at 13:49










  • I often use $1+1$ to avoid such ambiguity.
    – Derek Elkins
    Jan 4 at 18:30
















  • 1




    I've reversed Shaun's edit. I used unbolded $2$ on purpose for the discrete category with two objects, to distinguish it from $mathbf{2}$, which is the category with two objects and a unique morphism between them.
    – Clive Newstead
    Jan 4 at 13:49










  • I often use $1+1$ to avoid such ambiguity.
    – Derek Elkins
    Jan 4 at 18:30










1




1




I've reversed Shaun's edit. I used unbolded $2$ on purpose for the discrete category with two objects, to distinguish it from $mathbf{2}$, which is the category with two objects and a unique morphism between them.
– Clive Newstead
Jan 4 at 13:49




I've reversed Shaun's edit. I used unbolded $2$ on purpose for the discrete category with two objects, to distinguish it from $mathbf{2}$, which is the category with two objects and a unique morphism between them.
– Clive Newstead
Jan 4 at 13:49












I often use $1+1$ to avoid such ambiguity.
– Derek Elkins
Jan 4 at 18:30






I often use $1+1$ to avoid such ambiguity.
– Derek Elkins
Jan 4 at 18:30




















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