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Timeline for Injective objects in Mor(Ab)

Current License: CC BY-SA 3.0

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Oct 22, 2012 at 16:16 vote accept Martin Brandenburg
Oct 22, 2012 at 15:29 comment added Karol Szumiło It suffices to unravel the definitions of matching objects, in this case it is rather simple and indeed $A \to B \leftarrow C$ is injective if and only if both $A \to B$ and $C \to B$ are split surjective with injective sources.
Oct 22, 2012 at 15:25 comment added Martin Brandenburg Alright, thank you. So is there an easy way to describe the injectives in $\mathsf{Ab}^J$ for the $J$ above? Are these just those $(A \to B \leftarrow C)$ such that both $(A \to B)$ and $(B \leftarrow C)$ are injective in the category already described?
Oct 22, 2012 at 15:22 comment added Karol Szumiło Yes, it does. The easiest way to state the definition is that $J$ is inverse if there is functor $J^\mathrm{op} \to \mathbb{N}$ which reflects identities and your $J$ of course admits such a functor. (Sometimes it is useful to generalize by replacing $\mathbb{N}$ by some arbitrary ordinal).
Oct 22, 2012 at 15:14 comment added Martin Brandenburg Sorry for my ignorance, but does $J = (\bullet \rightarrow \bullet \leftarrow \bullet)$ fit into that framework?
Oct 22, 2012 at 15:12 comment added Karol Szumiło 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.
Oct 22, 2012 at 15:08 comment added Martin Brandenburg What do you mean by "$J$ is inverse"?
Oct 22, 2012 at 15:05 comment added Karol Szumiło 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.
Oct 22, 2012 at 14:56 comment added Martin Brandenburg 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.
Oct 22, 2012 at 12:17 history edited Karol Szumiło CC BY-SA 3.0
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Oct 22, 2012 at 12:11 history edited Karol Szumiło CC BY-SA 3.0
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Oct 22, 2012 at 11:44 history answered Karol Szumiło CC BY-SA 3.0