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# Exercise 93 - Chapter 3

edited June 2018

If $$D : \textbf{1} \rightarrow \textbf{Set}$$ is a functor, what is the limit of $$D$$? Compute it using Theorem 3.90, and check your answer against Definition 3.87.

Definition 3.87

Let $$D : \mathcal{J} \rightarrow \mathcal{C}$$ be a diagram. A cone $$(C, c_* )$$ over $$D$$ consists of

(i) an object $$C \in Ob(\mathcal{C}$$;

(ii) for each object $$j \in Ob(\mathcal{J})$$, a morphism $$c_j : C \rightarrow D(j)$$.

To be a cone, these must satisfy the following property:

(a) for each $$f : i \rightarrow j$$ in $$\mathcal{J}$$ , we have $$c_j = c_i . D( f )$$.

A morphism of cones $$(C, c_* ) \rightarrow (C' , c'_* )$$ is a morphism $$a : C \rightarrow C'$$ in $$\mathcal{C}$$ such that for all $$j \in \mathcal{J}$$ we have $$c_j = a.c'_j$$. Cones over $$D$$, and their morphisms, form a category $$\textbf{Cone}(D)$$.

The limit of $$D$$, denoted $$lim(D)$$, is the terminal object in the category $$\textbf{Cone}(D)$$. Say it is the cone $$lim(D) = (C, c_* )$$; we refer to $$C$$ as the limit object and the map $$c_j$$ for any $$j \in \mathcal{J}$$ as the $$j^{th}$$ projection map.

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1.

Via Theorem 3.90:

$$V=\{1\}$$. $$\mathrm{lim}_\textbf{1}D=\{(d_1)|d_1\in D(1)\,\mathrm{and}\,D(\mathrm{id}_1)(d_1)=d_1\}$$ together with a projection map $$p_1((d_1))=d_1$$. However, since $$D$$ is a functor, the constraint is always satisfied, and the projection map is the identity. So, the limit of $$D:\textbf{1}\to\textbf{Set}$$ is the set that it picks out in Set.

Via Definition 3.87:

Let $$S\in\textbf{Set}$$ be the set that $$D$$ picks out. A cone over $$D$$ consists of a set $$C$$ and a function $$f:C\to S$$ (the cone property is trivial here, just like the constraint in theorem 3.90). Morphisms between these cones are functions $$a:C\to C'$$ such that for $$g:C'\to S$$, $$g\circ a=f$$. A terminal object in this category Cone($$D$$) is a set $$T$$ and a function $$h:T\to S$$ such that for every set $$C$$ and function $$f:C\to S$$ there is a unique $$b:C\to T$$ with $$h\circ b=f$$. For each element $$s\in S$$ there is a cone $$\{ * \},f_s:\{ * \}\to S\}$$ where $$f_s(*)=s$$. Each $$b_s$$ such that $$h\circ b_s=f_s$$ must pick out an element in $$T$$, since the images of each $$f_s$$ are distinct, the images of each $$b_s$$ must also be distinct, so $$\lvert T\rvert\geq\lvert S\rvert$$ and $$h$$ must be surjective. Uniqueness of $$b_s$$ forces $$h$$ to be injective, so $$h$$ is an isomorphism and $$T\cong S$$. This shows that following defintion 3.87 agrees with following theorem 3.90.

Comment Source:Via Theorem 3.90: \$$V=\\{1\\}\$$. \$$\mathrm{lim}\_\textbf{1}D=\\{(d_1)|d\_1\in D(1)\,\mathrm{and}\,D(\mathrm{id}\_1)(d_1)=d_1\\}\$$ together with a projection map \$$p_1((d_1))=d_1\$$. However, since \$$D\$$ is a functor, the constraint is always satisfied, and the projection map is the identity. So, the limit of \$$D:\textbf{1}\to\textbf{Set}\$$ is the set that it picks out in **Set**. Via Definition 3.87: Let \$$S\in\textbf{Set}\$$ be the set that \$$D\$$ picks out. A cone over \$$D\$$ consists of a set \$$C\$$ and a function \$$f:C\to S\$$ (the cone property is trivial here, just like the constraint in theorem 3.90). Morphisms between these cones are functions \$$a:C\to C'\$$ such that for \$$g:C'\to S\$$, \$$g\circ a=f\$$. A terminal object in this category **Cone**(\$$D\$$) is a set \$$T\$$ and a function \$$h:T\to S\$$ such that for every set \$$C\$$ and function \$$f:C\to S\$$ there is a unique \$$b:C\to T\$$ with \$$h\circ b=f\$$. For each element \$$s\in S\$$ there is a cone \$$\\{ * \\},f\_s:\\{ * \\}\to S\\}\$$ where \$$f_s(*)=s\$$. Each \$$b_s\$$ such that \$$h\circ b_s=f_s\$$ must pick out an element in \$$T\$$, since the images of each \$$f_s\$$ are distinct, the images of each \$$b_s\$$ must also be distinct, so \$$\lvert T\rvert\geq\lvert S\rvert\$$ and \$$h\$$ must be surjective. Uniqueness of \$$b_s\$$ forces \$$h\$$ to be injective, so \$$h\$$ is an isomorphism and \$$T\cong S\$$. This shows that following defintion 3.87 agrees with following theorem 3.90.
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