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I am trying to show that the functor category $\operatorname{Func}(J,Ab)$ has enough injectives (meaning that for each $F\in \operatorname{Func}(J,Ab)$ there is an injective object $I\in \operatorname{Func}(J,Ab)$ and a monomorphism $F\to I$ in $\operatorname{Func}(J,Ab)$), where $Ab$ is the category of abelian groups and $J$ is the category

$$0\leftarrow 1 \leftarrow 2 \leftarrow \cdots$$

(so $J$ is the opposite of the category associated to the poset of non-negative integers).

($\operatorname{Func}(J,Ab)$ is the category of functors from $J$ to $Ab$, with arrows being natural transformations).

Here is what I was able to do:

I can show that $Ab$ has enough injectives.

I know that a natural transformation $\phi: F\to G$ is a monic arrow in $\operatorname{Func}(J,Ab)$ if and only if $\phi_i:F(i)\to G(i)$ is monic for all $i$.

Given $F\in \operatorname{Func}(J,Ab)$, I can find $I\in \operatorname{Func}(J,Ab)$ and a monomorphism $F\rightarrow I$, where $I(i)$ is injective for all $i$ (I just use the fact that $Ab$ has enough injectives to construct monomorphisms $\phi_i:F(i)\rightarrow I(i)$ with $I(i)$ injective and then use the fact that the $I(i)$ are injective to define $I$ on arrows so that $\phi$ becomes a natural transformation $F\to I$). The problem is that now I can't prove that $I$ is an injective object in $\operatorname{Func}(J,Ab)$:

  • Is it true that if all the $I(i)$ are injective in $Ab$ then $I$ is injective in $\operatorname{Func}(J,Ab)$?

    (I don't think so, but I didn't find any counterexamples).

I know that if I take an injective object $A\in Ab$ and consider the constant functor $F(i)=A$ for all $i$ and $F(\alpha)=\operatorname{id}_{A}$ for all arrows $\alpha$ in $J$, then $F$ is an injective object in $\operatorname{Func}(J,Ab)$.

I tried using limits:

I know that the functor $\lim:\operatorname{Func}(J,Ab)\to Ab$ is left exact and I used this to show that if $I\in \operatorname{Func}(J,Ab)$ is an injective object then $\lim I\in Ab$ is an injective object.

  • Is it true that if $\lim\;I$ is injective then $I$ is injective?

I would be very grateful for any hints.

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One way to do this is to use the general theorem proved by Grothendieck in his Tôhoku paper. This is explained in Weibel's book on homological algebra, for example. –  Mariano Suárez-Alvarez Feb 10 '12 at 21:30
    
I think you will find a proof in the first chapters of Gabriel's thesis (Categories abeliennes). –  Martin Brandenburg Mar 2 '12 at 8:50
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2 Answers 2

Here is a sledgehammer proof. (I have not tried to find an elementary proof.)

Let $\mathcal{C}$ be an arbitrary small category, and let $\hat{\mathcal{C}} = [\mathcal{C}^\textrm{op}, \textbf{Set}]$ be the presheaf topos on $\mathcal{C}$. The functor category $[\mathcal{C}^\textrm{op}, \textbf{Ab}]$ is equivalent to the category $\textbf{Ab}(\hat{\mathcal{C}})$ of all abelian group objects in $\hat{\mathcal{C}}$, and $\hat{\mathcal{C}}$ is a Grothendieck topos, so $\textbf{Ab}(\hat{\mathcal{C}})$ has enough injectives. (See Johnstone [Topos Theory, p. 261], for example.)

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Thank you for your answer! Unforntunately I do not know enough about category theory or topos theory to understand it. –  Manuel Araújo Jan 28 '12 at 16:06
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I'm pretty sure the answer to your first question is yes. Given the $I$ that you constructed, such that each $I(i)$ is injective, then the functor $I$ is injective. Suppose we have a monic arrow $f: G \hookrightarrow H$ of functors in $Func(J,Ab)$, and suppose we're given an arrow $g: G \rightarrow I$. Since each $I(i)$ is injective and each $f(i)$ is monic, by the universal property of injective objects there exist $h(i): H(i) \rightarrow I(i)$ extending $g(i)$, i.e. $h(i) \circ f(i)=g(i)$ for each $i$. These $h(i)$ define a functor $h:H \rightarrow I$ which extends $g$, i.e. $h \circ f=g$, so this verifies the universal property of an injective object.

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The problem I have is with showing that the h(i) define a natural transformation $h:H\to I$. In the meantime, my professor has written a sketch of solution, I will post the solution here as soon as possible. –  Manuel Araújo Feb 11 '12 at 1:36
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