I have been searching for the definition of a Corepresentable functor here https://ncatlab.org/nlab/show/small+object but still I did not get what exactly is its definition. I also love the definition of representable functor from here https://en.wikipedia.org/wiki/Representable_functor but still I do not exactly understand the definition of a Corepresentable functor. Could someone tell me the defintion please and how it differs from the definition of representable functor?
Corepresentable functor definition.
category-theorydefinitionrepresentable-functor
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The last sentence explains how the empty biproduct is a zero object. Namely the empty biproduct, I'll call it $0$, is an empty product, thus a terminal object, and an empty coproduct, thus an initial object. Since it is both a terminal and initial object, it is a zero object.
I'll note that there is actually a relevant error that might be confusing you, namely that the conditions on the morphisms only imply that the biproduct is both a product and coproduct in the case when the biproduct is not empty. If we leave out this part of the definition in the empty case, then the definition would read "a biproduct is any object together with a collection of no morphisms," which is not what we want.
Question 1. A typical example is the functor $\mathrm{Nil} : \mathbf{Ring} \to \mathbf{Set}$ of nilpotent elements in a ring. It is a subfunctor of the forgetful functor, which is represented by $\mathbb{Z}[T]$, but $\mathrm{Nil}$ is not representable: If $u$ is a universal element, then $u^n=0$ for some $n$, which would imply that every nilpotent $x$ in any ring would satisfy $x^n=0$, which is wrong. At least, $\mathrm{Nil}$ is ind-representable, since we have $$\mathrm{Nil} \cong \varinjlim_n \mathrm{Hom}(\mathbb{Z}[T]/T^n,-).$$ Similarly, the functor $\mathrm{Tors} : \mathbf{Grp} \to \mathbf{Set}$ of torsion elements in groups is not representable, but it is ind-representable since $$\mathrm{Tors} \cong \varinjlim_n \mathrm{Hom}(C_n,-).$$
Question 2. The following basic result is proved in EGA I. It can be used, for instance, to prove that fiber products of schemes exist. You can see it as a "gluing process guided by a functor".
Let $\mathbf{Sch}$ be the category of schemes (the statement works just as well for ringed spaces, locally ringed spaces, manifolds etc.). A functor $F : \mathbf{Sch}^{\mathrm{op}} \to \mathbf{Set}$ is represesentable if and only if the following two conditions hold:
- $F$ is a sheaf: For every open covering $U = \bigcup_i U_i$ of a scheme $U$ the diagram $F(U) \to \prod_i F(U_i) \rightrightarrows \prod_{i,j} F(U_i \cap U_j)$ is exact.
- $F$ has an open covering by representable functors.
Here, a subfunctor $G \subseteq F$ is called open when for every scheme $X$ and every element $a \in F(X)$, thus corresponding to a morphism $a : \mathrm{Hom}(-,X) \to F$, the pullback $G \times_F \mathrm{Hom}(-,X) \to \mathrm{Hom}(-,X)$ is isomorphic to $\mathrm{Hom}(-,U) \to \mathrm{Hom}(-,X)$ for some open subscheme $U \subseteq X$. Also, a family of subfunctors $G_i \subseteq F$ covers $F$ when for every scheme $X$ which is the spectrum of a field (not for any scheme!), we have $\bigcup_i G_i(X) = F(X)$.
Best Answer
A corepresentable is just $\mathrm{Hom}(c,-)$ as opposed to the representable $\mathrm{Hom}(-,c).$ Many people will call both representable.