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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.
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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.
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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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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 $M_n$'s 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 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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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 nonfull full subcategory $A^{(X)}$ of $A^X$ whose objects are the collections $M=(M_x)_{x\in X}$ such that there exists a small finite subset $E\subset X$ with the property that $M_x\simeq 0$ for $x\notin E$. Morphisms (which are not the identity) $M\to N$ are the collections of maps $M_x\to N_x$ which are zero for all but a finite number of indices. 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 $(M_n)_n=M$. Then the product of the $M_n$'s is not representable in $\widehat C$ (I leave this as an exercise: check that the product of the $M_n$'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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2
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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 nonfull subcategory $A^{(X)}$ of $A^X$ whose objects are the collections $M=(M_x)_{x\in X}$ such that there exists a small subset $E\subset X$ with the property that $M_x\simeq 0$ for $x\notin E$. Morphisms (which are not the identity) $M\to N$ are the collections of maps $M_x\to N_x$ which are zero for all but a finite number of indices. 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 $(M_n)_n=M$. Then the product of the $M_n$'s is not representable in $\widehat C$ (I leave this as an exercise: check that the product of the $M_n$'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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1
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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 nonfull subcategory $A^{(X)}$ of $A^X$ whose objects are the collections $M=(M_x)_{x\in X}$ such that there exists a small subset $E\subset X$ with the property that $M_x\simeq 0$ for $x\notin E$. Morphisms $M\to N$ are the collections of maps $M_x\to N_x$ which are zero for all but a finite number of indices. 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 $(M_n)_n=M$. Then the product of the $M_n$'s is not representable in $\widehat C$ (I leave this as an exercise: check that the product of the $M_n$'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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