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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*.

## Comments

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.`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.`