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Let $\mathcal{A}$ be an Abelian category with enough injectives. Is it always possible to make the injective embedding functorial? By this I mean that there should exist a functor $I \colon \mathcal{A} \to \mathcal{A}$ and a natural transformation $\operatorname{id} \to I$ such that for all objects $A$, the mapping $A \to I(A)$ is a monomorphism. This should be the same as this definition from the Stacks project. (edit: in a first version, I was requiring the functor to be additive, which is not what I had in mind even in the case of modules, as pointed out by Jeremy Rickard)

The Stacks project distinguishes categories with enough injectives from categories with functorial injective embeddings, so the two notions should be different. But I realized that I cannot think of an example of a category with enough injectives that does not admit functorial injective embeddings.

Motivation: when constructing derived functors, one has to choose injective resolutions, but such resolutions depend in general from arbitary choices. When functorial injective embeddings exist one can make an injective resolution functor, simplifying the construction of derived functors. The choices turn out not to matter much: for instance, every two injective resolutions are quasi-isomorphic, hence isomorphic in the derived category (say, bounded from the left). But it is easier to motivate the construction of derived functors first, and only later define the derived category.

The lack of functorial choices makes matters more complicated: for instance, Voisin, Hodge Theory and Complex Algebraic Geometry I, page 99, remarks explicitly that, for a left exact functor $F \colon \mathcal{A} \to \mathcal{A}'$, $R^{i}F$ is not a functor, since $R^{i}F(A)$ is only defined up to isomorphism in $\mathcal{A}'$. Many other books gloss on this point and state that since $R^{i}F(A)$ does not depend - up to isomorphism - from the injective resolution, the $R^{i}F$ are well-defined functors.

Let $\mathcal{A}$ be an Abelian category with enough injectives. Is it always possible to make the injective embedding functorial? By this I mean that there should exist a functor $I \colon \mathcal{A} \to \mathcal{A}$ and a natural transformation $\operatorname{id} \to I$ such that for all objects $A$, the mapping $A \to I(A)$ is a monomorphism. This should be the same as this definition from the Stacks project. (edit: in a first version, I was requiring the functor to be additive, which is not what I had in mind even in the case of modules, as pointed out by Jeremy Rickard)

The Stacks project distinguishes categories with enough injectives from categories with functorial injective embeddings, so the two notions should be different. But I realized that I cannot think of an example of a category with enough injectives that does not admit functorial injective embeddings.

edit removed a motivation comment that was sparking more discussion than necessary and distracting from the main question.

Let $\mathcal{A}$ be an Abelian category with enough injectives. Is it always possible to make the injective embedding functorial? By this I mean that there should exist an additive functor $I \colon \mathcal{A} \to \mathcal{A}$ and a natural transformation $\operatorname{id} \to I$ such that for all objects $A$, the mapping $A \to I(A)$ is a monomorphism. This should be the same as this definition from the Stacks project.

The Stacks project distinguishes categories with enough injectives from categories with functorial injective embeddings, so the two notions should be different. But I realized that I cannot think of an example of a category with enough injectives that does not admit functorial injective embeddings.

Motivation: when constructing derived functors, one has to choose injective resolutions, but such resolutions depend in general from arbitary choices. When functorial injective embeddings exist one can make an injective resolution functor, simplifying the construction of derived functors. The choices turn out not to matter much: for instance, every two injective resolutions are quasi-isomorphic, hence isomorphic in the derived category (say, bounded from the left). But it is easier to motivate the construction of derived functors first, and only later define the derived category.

The lack of functorial choices makes matters more complicated: for instance, Voisin, Hodge Theory and Complex Algebraic Geometry I, page 99, remarks explicitly that, for a left exact functor $F \colon \mathcal{A} \to \mathcal{A}'$, $R^{i}F$ is not a functor, since $R^{i}F(A)$ is only defined up to isomorphism in $\mathcal{A}'$. Many other books gloss on this point and state that since $R^{i}F(A)$ does not depend - up to isomorphism - from the injective resolution, the $R^{i}F$ are well-defined functors.

Let $\mathcal{A}$ be an Abelian category with enough injectives. Is it always possible to make the injective embedding functorial? By this I mean that there should exist a functor $I \colon \mathcal{A} \to \mathcal{A}$ and a natural transformation $\operatorname{id} \to I$ such that for all objects $A$, the mapping $A \to I(A)$ is a monomorphism. This should be the same as this definition from the Stacks project. (edit: in a first version, I was requiring the functor to be additive, which is not what I had in mind even in the case of modules, as pointed out by Jeremy Rickard)

The Stacks project distinguishes categories with enough injectives from categories with functorial injective embeddings, so the two notions should be different. But I realized that I cannot think of an example of a category with enough injectives that does not admit functorial injective embeddings.

Motivation: when constructing derived functors, one has to choose injective resolutions, but such resolutions depend in general from arbitary choices. When functorial injective embeddings exist one can make an injective resolution functor, simplifying the construction of derived functors. The choices turn out not to matter much: for instance, every two injective resolutions are quasi-isomorphic, hence isomorphic in the derived category (say, bounded from the left). But it is easier to motivate the construction of derived functors first, and only later define the derived category.

The lack of functorial choices makes matters more complicated: for instance, Voisin, Hodge Theory and Complex Algebraic Geometry I, page 99, remarks explicitly that, for a left exact functor $F \colon \mathcal{A} \to \mathcal{A}'$, $R^{i}F$ is not a functor, since $R^{i}F(A)$ is only defined up to isomorphism in $\mathcal{A}'$. Many other books gloss on this point and state that since $R^{i}F(A)$ does not depend - up to isomorphism - from the injective resolution, the $R^{i}F$ are well-defined functors.

Let $\mathcal{A}$ be an Abelian category with enough injectives. Is it always possible to make the injective embedding functorial? By this I mean that there should exist an additive functor $I \colon \mathcal{A} \to \mathcal{A}$ and a natural transformation $\operatorname{id} \to I$ such that for all objects $A$, the mapping $A \to I(A)$ is a monomorphism. This should be the same as this definition from the Stacks project.

The Stacks project distinguishes categories with enough injectives from categories with functorial injective embeddings, so the two notions should be different. But I realized that I cannot think of an example of a category with enough injectives that does not admit functorial injective embeddings.

Motivation: when constructing derived functors, one has to choose injective resolutions, but such resolutions depend in general from arbitary choices. When functorial injective embeddings exist one can make an injective resolution functor, simplifying the construction of derived functors. The choices turn out not to matter much: for instance, every two injective resolutions are quasi-isomorphic, hence isomorphic in the derived category (say, bounded from the left). But it is easier to motivate the construction of derived functors first, and only later define the derived category.

The lack of functorial choices makes matters more complicated: for instance, Voisin, Hodge Theory and Complex Algebraic Geometry I, page 99, remarks explicitly that, for a left exact functor $F \colon \mathcal{A} \to \mathcal{A}'$, $R^{i}F$ is not a functor, since $R^{i}F(A)$ is only defined up to isomorphism in $\mathcal{A}'$. Many other books gloss on this point are state that since $R^{i}F(A)$ does not depend - up to isomorphism - from the injective resolution, the $R^{i}F$ are well-defined functors.

Let $\mathcal{A}$ be an Abelian category with enough injectives. Is it always possible to make the injective embedding functorial? By this I mean that there should exist an additive functor $I \colon \mathcal{A} \to \mathcal{A}$ and a natural transformation $\operatorname{id} \to I$ such that for all objects $A$, the mapping $A \to I(A)$ is a monomorphism. This should be the same as this definition from the Stacks project.

The Stacks project distinguishes categories with enough injectives from categories with functorial injective embeddings, so the two notions should be different. But I realized that I cannot think of an example of a category with enough injectives that does not admit functorial injective embeddings.

Motivation: when constructing derived functors, one has to choose injective resolutions, but such resolutions depend in general from arbitary choices. When functorial injective embeddings exist one can make an injective resolution functor, simplifying the construction of derived functors. The choices turn out not to matter much: for instance, every two injective resolutions are quasi-isomorphic, hence isomorphic in the derived category (say, bounded from the left). But it is easier to motivate the construction of derived functors first, and only later define the derived category.

The lack of functorial choices makes matters more complicated: for instance, Voisin, Hodge Theory and Complex Algebraic Geometry I, page 99, remarks explicitly that, for a left exact functor $F \colon \mathcal{A} \to \mathcal{A}'$, $R^{i}F$ is not a functor, since $R^{i}F(A)$ is only defined up to isomorphism in $\mathcal{A}'$. Many other books gloss on this point and state that since $R^{i}F(A)$ does not depend - up to isomorphism - from the injective resolution, the $R^{i}F$ are well-defined functors.