The following lemma is often useful to prove the existence of universal objects in big categories, please see the discussion in Remark 4.25.2.
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
F commutes with limits,
there exist 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 exist 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{\mathrm{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{\mathrm{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{\mathrm{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
Theorem 4.25.3 (Adjoint functor theorem). Let G : \mathcal{C} \to \mathcal{D} be a functor of big categories. Assume \mathcal{C} has limits, G commutes with them, and for every object y of \mathcal{D} there exists a set of pairs (x_ i, f_ i)_{i \in I} with x_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), f_ i \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(y, G(x_ i)) such that for any pair (x, f) with x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), f \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(y, G(x)) there are an i and a morphism h : x_ i \to x such that f = G(h) \circ f_ i. Then G has a left adjoint F.
Proof.
The assumptions imply that for every object y of \mathcal{D} the functor x \mapsto \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(y, G(x)) satisfies the assumptions of Lemma 4.25.1. Thus it is representable by an object, let's call it F(y). An application of Yoneda's lemma (Lemma 4.3.5) turns the rule y \mapsto F(y) into a functor which by construction is an adjoint to G. We omit the details.
\square
Comments (2)
Comment #6028 by Rex on
Comment #6175 by Johan on