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Consider the abelian (Grothendieck) category $\mathcal{C} := \mathrm{Fun}(\{0<1\},\mathrm{Ab}) = \mathrm{Mor}(\mathrm{Ab})$. Objects are morphisms $(A \to B)$ of abelian groups, morphisms are commutative diagrams. Equivalently, this is the category of abelian sheaves on the Sierpinski space.

Question. How do injective objects in $\mathcal{C}$ look like?

Since injective sheaves are stable under restriction (use extension by zero), clearly $(A \to B)$ injective implies that $A$ is injective. But is this sufficient (probably not)? When $A,B$ are injective, is the same true for $(A \to B)$?

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Morphisms of abelian groups correspond to maximally symmetric 2-groups, but the ordinary morphisms you have (commuting squares) aren't all the morphisms. It would be interesting to see the answer to this question in the case when the quasi-isomorphisms are formally inverted in a 2-cateorical sense (e.g. butterflies a la Noohi et al). –  David Roberts Oct 22 '12 at 8:28
Split surjections with injective source. –  Fernando Muro Oct 22 '12 at 9:35
I can verify that split surjections with injective source represent injective objects in $C$, but what about the other direction? –  Martin Brandenburg Oct 22 '12 at 10:41
I believe $\mathcal C$ is equivalent to the module category of the triangular matrix ring $\begin{pmatrix} \mathbb Z & 0\\ \mathbb Z & \mathbb Z \end{pmatrix}$ –  Dag Oskar Madsen Oct 22 '12 at 11:45

1 Answer 1

up vote 3 down vote accepted

I will use notation $A_0 \to A_1$ for objects of $\mathrm{Mor}(\mathrm{Ab})$.

EDIT: previously I claimed something stronger (that I can produce lifting properties in the functor category without factorizations), but I am not so sure about it.

The following is a lot more general than necessary, but I think this added generality is also useful. Let $(\mathcal{L}, \mathcal{R})$ be a weak factorization system in a category $\mathcal{C}$ with enough colimits and limits for the following to make sense. Let $J$ be a Reedy category. Then in the functor category $\mathcal{C}^J$ the "Reedy $\mathcal{L}$-cofibrations" and "Reedy $\mathcal{R}$-fibrations" form a weak factorization system. By "Reedy $\mathcal{L}$-cofibrations" I mean morphisms of diagrams $X \to Y$ such that for every $j \in J$ the latching morphism $X_j \sqcup_{L_j X} L_j Y \to Y_j$ is in $\mathcal{L}$ and dually "Reedy $\mathcal{R}$-fibrations" are morphisms $X \to Y$ such that for every $j \in J$ the matching morphism $X_j \to M_j X \times_{M_j Y} Y_j$ is in $\mathcal{R}$. The proof is exactly as in the construction of the Reedy model structures and can be found for example in Hovey's Model Categories.

Now we take $\mathcal{C} = \mathrm{Ab}$, $\mathcal{L} = $ monomorphisms and $J = [1]$. Then $\mathcal{R}$ are split epimorphisms with injective kernel. The lifting properties are easily verified while the factorizations use the fact that there are enough injectives in $\mathrm{Ab}$. If $f : A \to B$ is a map in $\mathrm{Ab}$, pick an injective hull $i : A \to \hat A$, then $f$ factors as an injection $[i, f] : A \to \hat A \oplus B$ followed by a split surjection with injective kernel $\hat A \oplus B \to B$. We consider $J$ as a Reedy category where $0$ has degree $1$ and $1$ has degree $0$. Then "Reedy $\mathcal{L}$-cofibrations" are monomorphisms again, so an object $X$ is injective if and only if the map $X \to 0$ is a "Reedy $\mathcal{R}$-fibration" i.e. when both $X_1 \to 0$ and $X_0 \to X_1$ are split epimorphisms with injective kernel i.e. when $X_0 \to X_1$ is a split epimorphism with injective source.

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Thank you! So here is a direct proof: Let $(A \to B)$ be an injective object. Then $A$ is injective. Since $(A \to B)$ embeds into $(A \oplus B \to B)$, there is a retraction $(A \oplus B \to B) \to (A \to B)$. This shows that $A\to B$ is split. –  Martin Brandenburg Oct 22 '12 at 14:56
Yes, I admit that my proof is an overkill. Especially that it relies on the extra assumption that there are enough injective objects. But it really is more general, it gives characterization of injectives in $\mathcal{A}^J$ where $\mathcal{A}$ is any abelian category with enough injectives and $J$ is inverse. –  Karol Szumiło Oct 22 '12 at 15:05
What do you mean by "$J$ is inverse"? –  Martin Brandenburg Oct 22 '12 at 15:08
Every Reedy category has a direct part (morphisms that raise the degree) and an inverse part (those that lower the degree). An inverse category is a Reedy category with trivial direct part so that all non-identity morphisms lower the degree. This assumption was used to conclude that "Reedy $\mathcal{L}$-cofibrations" are just levelwise monomorphisms i.e. monomorphisms. –  Karol Szumiło Oct 22 '12 at 15:12
Sorry for my ignorance, but does $J = (\bullet \rightarrow \bullet \leftarrow \bullet)$ fit into that framework? –  Martin Brandenburg Oct 22 '12 at 15:14

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