Would it be possible to enlighten me (or even better give a reference) about enough projectives (injectives) in functor categories?

Here is a precise question. Let $C$ be a small category, whose total morphism set has cardinality $\alpha$. Let $A$ be an abelian category with enough projectives (dually, injectives) and coproducts (products) up to cardinality $\alpha$. The functor category $A^C$ is clearly abelian.

Question Does $A^C$ have enough projectives (dually, injectives)?

A reference would be ideal but an explanation would be very welcome too. I am getting fricasseeed this morning by this question: surely, the issue should be whether one can cover an object in $A$ by a projective object functorially but I have no idea how to spell it out...

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    $\begingroup$ Even if you could choose a functorial projective cover in $\mathcal{A}$, there is still the fact that diagrams that are componentwise projective need not be projective. For instance, $\mathcal{C}$ could be the category freely generated by one endomorphism and $\mathcal{A}$ could be the category of $k$-vector spaces; then a diagram $\mathcal{C} \to \mathcal{A}$ is the same thing as a $k [x]$-module. $\endgroup$
    – Zhen Lin
    Apr 8, 2014 at 13:02
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    $\begingroup$ @ZhenLin Moreover: If one chooses $\mathcal{A}$ to be the category of finite dimensional $k$-vector spaces, then $\mathcal{A}$ has enough projective for the cardinality of $Ob(\mathcal{C})=\{\ast\}$ but there are no finite dimensional $k[x]$-modules. $\endgroup$ Apr 8, 2014 at 13:07
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    $\begingroup$ If $A$ were a Grothendieck category the answer would be yes, even under smaller assumptions. I wonder what you have in mind when you require $A$ to have just 'very small' (co)products. $\endgroup$ Apr 8, 2014 at 13:10
  • $\begingroup$ Good point and a counter-example too: take your $C$ and $A$ the category of finite dimensional vector spaces... I am editing the question to address this. $\endgroup$
    – Bugs Bunny
    Apr 8, 2014 at 13:11
  • $\begingroup$ @ Fernando: what is the reference for this? I am kind of see why this helps with injectives (probably, $A^C$ will be Grothendieck too, will it not?) but I am not sure about projectives... $\endgroup$
    – Bugs Bunny
    Apr 8, 2014 at 13:16

1 Answer 1


For each object $c$ in $\mathcal{C}$, let $c^* : [\mathcal{C}, \mathcal{A}] \to \mathcal{A}$ be evaluation at $c$. It is an exact functor, so if a left adjoint $c_! : \mathcal{A} \to [\mathcal{C}, \mathcal{A}]$ exists, $c_!$ will preserve projective objects. Assume $\mathcal{C}$ has $\le \alpha$ morphisms and $\mathcal{A}$ has coproducts for families of $\le \alpha$ objects. Then the left adjoint $c_!$ exists and can be computed the following formula: $$(c_! A) (c') = \mathcal{C} (c, c') \odot A$$ (Here, $X \odot A$ denotes the coproduct of $X$-many copies of $A$.)

Now, let $F$ be an object in $[\mathcal{C}, \mathcal{A}]$. For each object $c$ in $\mathcal{C}$, choose a projective cover $P_c \twoheadrightarrow F c$. By adjointness, we obtain morphisms $c_! P_c \to F$ in $\mathcal{A}$; note that the composite $P_c \to c^* c_! P_c \to F c$ is the epimorphism we started with, so $c^* c_! P_c \to F$ is an epimorphism in particular. Now, form the object $P = \bigoplus_{c \in \operatorname{ob} \mathcal{C}} c_! P_c$; then there is a morphism $P \to F$ such that the components $P c \to F c$ are all epimorphisms. Furthermore, each $c_! P_c$ is projective, and the class of projective objects is closed under coproducts, so we have obtained the required projective cover of $F$.

I'm afraid I do not have a reference, but the above is essentially the same as the proof that (say) $[\mathcal{C}, \mathbf{Ab}]$ has enough projectives. It can be further generalised to the case where $\mathcal{C}$ is preadditive and $[\mathcal{C}, \mathcal{A}]$ is the category of additive functors.


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