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Let $I$ be a small category and suppose $F:C\rightarrow Set$ and $G:I\rightarrow C$ are functors.

(i) How to construct a morphism $\alpha:F(\varprojlim_{i}G(i))\rightarrow \varprojlim_{i}F(G(i))$ using the projections $\varprojlim G(i)\rightarrow G(i)$.

(ii) How to prove that if $F$ is representable then $\alpha$ is an isomorphism?

(iii) What will be the corresponding statement if $F$ is a functor $F:C^{op}\rightarrow Set$?

Thank you for your help and time.

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I will write $\lim$ for $\varprojlim$, and $\lim G$ for the limit of the diagram $G : I \to C$.

For (i), note that by definition, there are maps (I wouldn't call them "projections") $\lim G \to G(i)$ for each $i\in I$, and moreover that these maps are natural in $i$, so that for each arrow $i \to j$, we get a commutative triangle. Applying $F$ gives a system of maps $F(\lim G) \to (F\circ G)(i)$, which is also natural in $i$. The desired morphism $\alpha$ is then unique determined by the universal property of $\lim (F\circ G)$.

Suppose now that $F$ is corepresentable, meaning that it is naturally isomoprhism to $\hom_C(f,-)$ for some object $f\in C$. (A better notation is to say that the functor $\hom_C(f,-)$ is corepresented by $f$, so that the contravariant functor $\hom_C(-,f)$ is the representable functor represented by $f$.) The universal property of the limit implies then that $\hom_C(f,\lim G) = $ the set of systems of morphisms $f \to G(i)$ that are natural in $i$; but such a system is exactly a system of elements in $\hom_C(f,G(i))$, which is to say an element of $\lim \hom_C(f,G)$. The corollary is that corepresentable functors distribute over (i.e. commute with) limits. It is worth thinking about this until it becomes trivial.

Finally, limits in $C^{\mathrm{op}}$ are the same as colimits in $C$. Thus for any contravariant functor $F$, there is a canonical map $F \operatorname{colim} G \to \lim FG$, which is an isomorphism if $F$ is representable.

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  • $\begingroup$ Corepresentability is something else (at least in the theory of moduli spaces), you just mean representable. In some sense, every functor is covariant. $\endgroup$ – Martin Brandenburg Dec 3 '13 at 14:24
  • $\begingroup$ @MartinBrandenburg Oh, what else does corepresentable mean? I'm not well versed in moduli spaces. For many applications, (pre)sheaves are more basic than "functors to SET", and the representable ones are the most basic of all. Cosheaves are also basic, but less so. In the world of generalized manifolds, there are uses for both sheaves and cosheaves, and it can be important to distinguish whether an object is defined by its maps from or its maps to; hence for a classical manifold, whether you want to think of its representing or corepresenting functor. $\endgroup$ – Theo Johnson-Freyd Dec 4 '13 at 1:25
  • $\begingroup$ A set-valued functor $F$ is corepresentable if there is an initial representable functor $H$ with a morphism $F \to H$ (not vice versa which would lead to representability) and usually in moduli spaces one demands that $F(k) \to H(k)$ is an isomorphism for all fields $k$. $\endgroup$ – Martin Brandenburg Dec 5 '13 at 0:58

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