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I will use two constructions to get to an example of a cocomplete abelian category which is

My previous answer was not complete.

First, here is a rather systematic way to turn a small abelian category into a huge (but still locally small) one. Start with an additive category $A$ (that is a category enriched in abelian groups with finite direct sums), and choose huge set $X$ (or a class, if you prefer), say the one whose elements are the small sets. We can form the product category $A^X$ (which won't be locally small anymore, even if $A$ had this property), and then consider the full subcategory $A^{(X)}$ of $A^X$ whose objects are the collections $M=(M_x)_{x\in X}$ such that there exists a finite subset $E\subset X$ with the property that $M_x\simeq 0$ for $x\notin E$. The category $A^{(X)}$ is locally small, and, for any $x\in X$, we have an evaluation at $x$ functor$$ev_x:A^{(X)}\to A .$$It is easy to see that, whenever $A$ has some type of finite (co)limit, then so does $A^{(X)}$, and that the functors $ev_x$ commute with them. As the functors $ev_x$ form a conservative family, we see that, if $A$ is abeliancorrect indeed, then so is $A^{(X)}$. It is easy to see that , if $A$ is not equivalent to the zero category, then $A^{(X)}$ is not essentially small. For instance, if $A$ what remains is just the abelian category of finitely generated abelian groups, then $A^{(X)}$ is an abelian category which is not essentially small and which is not complete nor cocompletefollowing comment.

Second, there

There is a way to consider the free colimit completion in the setting of additive categories; I will essentially use construtions and results which can found in this article of B. Day and S. Lack (in the particular case of additive categories): arXiv:math/0610439. Let $C$ be a locally small additive category. I will write $\widehat C$ for the free completion of $C$ by small colimits (in the enriched sense). An explicit construction of $\widehat C$ is the following: if $Ab$ denotes the symmetric monoidal category of abelian groups, $\widehat C$ is the full subcategory of the category of additive functorswith $F''$ and $F'$ isomorphic to small sums of representable presheaves. The category $\widehat C$ is always locally small and cocomplete (and the Yoneda embedding $C\to\widehat C$ is the universal additive functor from $C$ to a cocomplete additive category. I claim that, if $C$ is abelian, then $\widehat C$ is abelian as well: indeed, it has finite limits (see Prop.4.4 in Day & Lack's paper) and the Yoneda embedding $C\to\widehat C$ commutes with limits; moreover, for any object $X$ of $C$ the evaluation at $X$ functor has both a left and a right adjoint (I leave as an exercise their explicit description) and thus is exact; as these evaluation functors form a conservative family, one deduces that $\widehat C$ is abelian whenever $C$ is abelian (in fact, it is sufficient for this that $C$ is finitely complete). Note also that, whenever they exist, limits of $\widehat C$ can be computed termwise. In particular, to check that a limit is not representable in $\widehat C$, it is sufficient to check that the corresponding presheaf over $C$ is not small.

Finally, let $C=A^{(X)}$ with $A$ any locally smallabelian category.I claim that, if $A$ is not equivalent to zero, then the cocomplete

The abelian category $\widehat C$ is not complete. To be more precise, consider a non zero object $M$ of $A$ as well an infinite subset $N\subset X$, and, for each $n\in N$, let $M_n$ known to be the object of $C$ defined by $(M_n)_x=0$ complete if $x\neq n$ and $(M_n)_n=M$. Given a finite subset $E$ one of $N$, write $M^E$ for the direct product of $M_n$ with $n$ running over $E$ (which following conditions is representable in $C$). This defines a functor from the opposite of the partially ordered set of non empty finite subsets $E$ of $N$ to $C$. Let $F$ be the limit of this diagram in the category of presheaves over satisfied: $C$. One can compute thatC$is essentialy small, for any object$N$of$C$, or$F(N)$C$ is the product of the abelian groups $Hom(N_x,M)$ for $x\in N$itself complete. I claim that Furthermore, if ever $F$ C$is not small. To see thiscocomplete, consider a small set$I$, a family$N_i$of objects of$C$, as well as a family of morphisms of presheaves$u_i:N_i\to F$. We may assume that each$N_i$then it is supported on a single element refexive subcategory of$X$. In particular\widehat C$, we have a map $v:I\to X$ such so that $N_i$ is zero outside the completeness of $v(i)$. For each \widehat C$implies the same property for$i\in I$C$. In conclusion, if $v(i)\in N$, ever there is a natural map $M_{v(i)}\to F$ induced by the identity an example of $M$. Let us define $M_x=0$ for $x\notin N$a cocomplete abelian category which is not complete, and consider the direct sum $G$ there must be one of all the presheaves form $M_{v(i)}$ \widehat C$for$i\in I$. We then have a natural map an additive category$G\to F$, C$ which is not essentially small and the map from the direct sum of the $N_i$'s factors through $G$. Therefore, it which is sufficient to check that the map $G\to F$ cannot be an epimorphism of presheavesnot complete. But the image of the map $G\to F$ Furthermore, if there is the direct sum $H$ a counter example, there must exist a small family of the representable presheaves $M_x$ for $x\in N$. For an object $P$ of $C$, $H(P)$ is thus the direct sum of the abelian groups $Hom(P_x,M)$ (with $x\in N$), while over $F(P)$ is their C$whose product . The fact that infinite direct sums do not coincide with infinite products in the category of abelian groups achieves the proofpreshaves is not small. So far, I did not succeed to find such a thing. 6 I modified again the very end of the argument: there is no exercise left to the reader anymore. Finally, let$C=A^{(X)}$with$A$any locally small abelian category. I claim that, if$A$is not equivalent to zero, then the cocomplete abelian category$\widehat C$is not complete. To be more precise, consider a non zero object$M$of$A$as well an infinite subset$N\subset X$, and, for each$n\in N$, let$M_n$be the object of$C$defined by$(M_n)_x=0$if$x\neq n$and$(M_n)_n=M$. Given a finite subset$E$of$N$, write$M^E$for the direct product of$M_n$with$n$running over$E$(which is representable in$C$). Then This defines a functor from the product opposite of the$M^E$'s (where partially ordered set of non empty finite subsets$E$run over all of$N$to$C$. Let$F$be the finite subsets limit of this diagram in the category of presheaves over$N$) C$. One can compute that, for any object $N$ of $C$, $F(N)$ is not representable in the product of the abelian groups $\widehat C$ (Hom(N_x,M)$for$x\in N$. I leave claim that$F$is not small. To see this, consider a small set$I$, a family$N_i$of objects of$C$, as an exercise: check well as a family of morphisms of presheaves$u_i:N_i\to F$. We may assume that the product each$N_i$is supported on a single element of$X$. In particular, we have a map$v:I\to X$such that$N_i$is zero outside$v(i)$. For each$i\in I$, if$v(i)\in N$, there is a natural map$M_{v(i)}\to F$induced by the identity of$M^E$'s in M$. Let us define $M_x=0$ for $x\notin N$, and consider the category direct sum $G$ of all the presheaves over $C$ cannot be M_{v(i)}$for$i\in I$. We then have a natural map$G\to F$, and the target map from the direct sum of the$N_i$'s factors through$G$. Therefore, it is sufficient to check that the map$G\to F$cannot be an epimorphism from a small of presheaves. But the image of the map$G\to F$is the direct sum$H$of the representable presheaves , and$M_x$for$x\in N$. For an object$P$of$C$,$H(P)$is thus the direct sum of the abelian groups$Hom(P_x,M)$(with$x\in N$), while$F(P)$is their product. The fact that infinite direct sums do not small)coincide with infinite products in the category of abelian groups achieves the proof. 5 edited body I will use two constructions to get to an example of a cocomplete abelian category which is not complete. First, here is a rather systematic way to turn a small abelian category into a huge (but still locally small) one. Start with an additive category$A$(that is a category enriched in abelian groups with finite direct sums), and choose huge set$X$(or a class, if you prefer), say the one whose elements are the small sets. We can form the product category$A^X$(which won't be locally small anymore, even if$A$had this property), and then consider the full subcategory$A^{(X)}$of$A^X$whose objects are the collections$M=(M_x)_{x\in X}$such that there exists a finite subset$E\subset X$with the property that$M_x\simeq 0$for$x\notin E$. The category$A^{(X)}$is locally small, and, for any$x\in X$, we have an evaluation at$x$functor $$ev_x:A^{(X)}\to A .$$ It is easy to see that, whenever$A$has some type of finite (co)limit, then so does$A^{(X)}$, and that the functors$ev_x$commute with them. As the functors$ev_x$form a conservative family, we see that, if$A$is abelian, then so is$A^{(X)}$. It is easy to see that, if$A$is not equivalent to the zero category, then$A^{(X)}$is not essentially small. For instance, if$A$is the abelian category of finitely generated abelian groups, then$A^{(X)}$is an abelian category which is not essentially small and which is not complete nor cocomplete. Second, there is a way to consider the free colimit completion in the setting of additive categories; I will essentially use construtions and results which can found in this article of B. Day and S. Lack (in the particular case of additive categories): arXiv:math/0610439. Let$C$be a locally small additive category. I will write$\widehat C$for the free completion of$C$by small colimits (in the enriched sense). An explicit construction of$\widehat C$is the following: if$Ab$denotes the symmetric monoidal category of abelian groups,$\widehat C$is the full subcategory of the category of additive functors $$F: C^{op}\to Ab$$ which consists of those$F$'s which are small, i.e. such that there exists a short exact sequence of the form $$F''\to F'\to F \to 0$$ with$F''$and$F'$isomorphic to small sums of representable presheaves. The category$\widehat C$is always locally small and cocomplete (and the Yoneda embedding$C\to\widehat C$is the universal additive functor from$C$to a cocomplete additive category. I claim that, if$C$is abelian, then$\widehat C$is abelian as well: indeed, it has finite limits (see Prop.4.4 in Day & Lack's paper) and the Yoneda embedding$C\to\widehat C$commutes with limits; moreover, for any object$X$of$C$the evaluation at$X$functor has both a left and a right adjoint (I leave as an exercise their explicit description) and thus is exact; as these evaluation functors form a conservative family, one deduces that$\widehat C$is abelian whenever$C$is abelian (in fact, it is sufficient for this that$C$is finitely complete). Note also that, whenever they exist, limits of$\widehat C$can be computed termwise. In particular, to check that a limit is not representable in$\widehat C$, it is sufficient to check that the corresponding presheaf over$C$is not small. Finally, let$C=A^{(X)}$with$A$any locally small abelian category. I claim that, if$A$is not equivalent to zero, then the cocomplete abelian category$\widehat C$is not complete. To be more precise, consider a non zero object$M$of$A$as well an infinite subset$N\subset X$, and, for each$n\in N$, let$M_n$be the object of$C$defined by$(M_n)_x=0$if$x\neq n$and$(M_n)_n=M$. Given a finite subset$E$of$N$, write$M^E$for the direct product of$M_n$with$n$running over$E$(which is representable in$C$). Then the product of the$M^E$'s (where$E$run over all the finite subsets of$N$) is not representable in$\widehat C$(I leave this as an exercise: check that the product of the$M_n$'s M^E$'s in the category of presheaves over $C$ cannot be the target of an epimorphism from a small sum of representable presheaves, and thus is not small).

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