Lemma 4.25.1. Let $\mathcal{C}$ be a big1 category which has limits. Let $F : \mathcal{C} \to \textit{Sets}$ be a functor. Assume that

1. $F$ commutes with limits,

2. there exists a family $\{ x_ i\} _{i \in I}$ of objects of $\mathcal{C}$ and for each $i \in I$ an element $f_ i \in F(x_ i)$ such that for $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $g \in F(y)$ there exists an $i$ and a morphism $\varphi : x_ i \to y$ with $F(\varphi )(f_ i) = g$.

Then $F$ is representable, i.e., there exists an object $x$ of $\mathcal{C}$ such that

$F(y) = \mathop{Mor}\nolimits _\mathcal {C}(x, y)$

functorially in $y$.

Proof. Let $\mathcal{I}$ be the category whose objects are the pairs $(x_ i, f_ i)$ and whose morphisms $(x_ i, f_ i) \to (x_{i'}, f_{i'})$ are maps $\varphi : x_ i \to x_{i'}$ in $\mathcal{C}$ such that $F(\varphi )(f_ i) = f_{i'}$. Set

$x = \mathop{\mathrm{lim}}\nolimits _{(x_ i, f_ i) \in \mathcal{I}} x_ i$

(this will not be the $x$ we are looking for, see below). The limit exists by assumption. As $F$ commutes with limits we have

$F(x) = \mathop{\mathrm{lim}}\nolimits _{(x_ i, f_ i) \in \mathcal{I}} F(x_ i).$

Hence there is a universal element $f \in F(x)$ which maps to $f_ i \in F(x_ i)$ under $F$ applied to the projection map $x \to x_ i$. Using $f$ we obtain a transformation of functors

$\xi : \mathop{Mor}\nolimits _\mathcal {C}(x, - ) \longrightarrow F(-)$

see Section 4.3. Let $y$ be an arbitrary object of $\mathcal{C}$ and let $g \in F(y)$. Choose $x_ i \to y$ such that $f_ i$ maps to $g$ which is possible by assumption. Then $F$ applied to the maps

$x \longrightarrow x_ i \longrightarrow y$

(the first being the projection map of the limit defining $x$) sends $f$ to $g$. Hence the transformation $\xi$ is surjective.

In order to find the object representing $F$ we let $e : x' \to x$ be the equalizer of all self maps $\varphi : x \to x$ with $F(\varphi )(f) = f$. Since $F$ commutes with limits, it commutes with equalizers, and we see there exists an $f' \in F(x')$ mapping to $f$ in $F(x)$. Since $\xi$ is surjective and since $f'$ maps to $f$ we see that also $\xi ' : \mathop{Mor}\nolimits _\mathcal {C}(x', -) \to F(-)$ is surjective. Finally, suppose that $a, b : x' \to y$ are two maps such that $F(a)(f') = F(b)(f')$. We have to show $a = b$. Consider the equalizer $e' : x'' \to x'$. Again we find $f'' \in F(x'')$ mapping to $f'$. Choose a map $\psi : x \to x''$ such that $F(\psi )(f) = f''$. Then we see that $e \circ e' \circ \psi : x \to x$ is a morphism with $F(e \circ e' \circ \psi )(f) = f$. Hence $e \circ e' \circ \psi \circ e = e$. Since $e$ is a monomorphism, this implies that $e'$ is an epimorphism, thus $a = b$ as desired. $\square$

[1] See Remark 4.2.2.

Comment #2300 by Marco D'Addezio on

There is a typo in the last sentence, $e'\circ e$ should be $e\circ e'$. Actually I would change the entire sentence with the following: "Since $e$ is a monomorphism, this implies that $e'$ is an epimorphism, thus $a=b$ as desired."

Comment #3328 by Kazuki Masugi on

Typo in the proof, “two maps such that F(a)(f)=F(b)(f).” should be “two maps such that F(a)(f’)=F(b)(f’).”

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