Timeline for If an abelian category $\mathcal{A}$ has enough injectives then so does $\mathrm{Ch}^{\geq 0}(\mathcal{A})$
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2014 at 11:10 | comment | added | W.Z. | Note: The mistake pointed out by J. Rickard is corrected in this answer body. | |
| Jul 31, 2014 at 10:59 | comment | added | W.Z. | To be clear, in $\mathrm{Ch}^{\geq 0}(\mathcal{A})$, if $\mathrm{H}^0(A^\cdot)$ is injective and $0\longrightarrow\mathrm{H}^0(A^\cdot)\longrightarrow A^0\longrightarrow A^1\longrightarrow\cdot\cdot\cdot$ is exact then $0\longrightarrow\mathrm{H}^0(A^\cdot)\longrightarrow A^0\longrightarrow\mathrm{ker}d^1\longrightarrow 0$ is split exact. Since we require $A^n$'s to be injective, so is $\mathrm{ker}d^1$. By induction, we have all injective $\mathrm{ker}d^n$'s. | |
| Jul 31, 2014 at 10:53 | comment | added | W.Z. | Yeah, I think you're right, we need each $\mathrm{ker}d^i$ to be injective or for the case of $\mathrm{Ch}^{\geq 0}(\mathcal{A})$ it is required to have $\mathrm{H}^0(A^\cdot)$ to be injective. Actually I introduced this mistaken condition but somehow it was credited as Community. Apology for this mathematical mistake! | |
| Jul 31, 2014 at 8:27 | comment | added | Jeremy Rickard | It's not true in general that an object of $\mathrm{Ch}(\mathcal{A})$ is injective if it's exact and each term is injective (which seems to have been introduced into your post in an edit by somebody else): you need $\operatorname{ker}(d^i)$ to be a direct factor of the degree $A^i$ (which is not automatic for complexes unbounded to the left). | |
| Jul 31, 2014 at 8:21 | comment | added | Jeremy Rickard | In my answer, the infinite products were of families of complexes such that in each degree only finitely many of the complexes are non-zero, which doesn't need AB3*. | |
| Jul 31, 2014 at 2:23 | comment | added | W.Z. | In Jeremy's answer, the embedding is $(\eta^n,f^n\circ\eta^n)^T:X^n\to I^n\oplus I^{n+1}$ where $\eta^n$ is the embedding $X^n\to I^n$ and $f^n:I^n\to I^{n+1}$ is an extension using injectivity of $I^{n+1}$. Might we need not (co)product of infinitely many complexes which imposes (AB3)* or (AB3) on $\mathcal{A}$, but only the criterion for injectivity in $\mathrm{Ch}^{\geq 0}(\mathcal{A})$ given in the OP? E.g., for $\mathrm{Ch}(\mathcal{A})$, $A^\cdot$ is injective iff $A^\cdot$ is exact and $A^n$'s and $\ker d^n$'s are injective. So the construction would apply to $\mathrm{Ch}(\mathcal{A})$. | |
| Jul 30, 2014 at 15:04 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |