Timeline for Splitting in additive categories
Current License: CC BY-SA 4.0
13 events
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| Aug 29 at 10:00 | history | edited | user267839 | CC BY-SA 4.0 |
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| Aug 19 at 9:10 | comment | added | Jeremy Rickard | In a general additive category, having a splitting map is not enough to ensure being the inclusion of a direct summand. E.g., in the additive category of real vector spaces with dimension not equal to one, an injective map $\mathbb{R}^2\to\mathbb{R}^3$ has a splitting map, but $\mathbb{R}^2$ is not a direct summand of $\mathbb{R}^3$ | |
| Aug 19 at 9:07 | comment | added | Jeremy Rickard | In a triangulated category, if $A\to B$ has a splitting map (i.e., a map $B\to A$ such that the composition $A\to B\to A$ is $\operatorname{id}_A$), then it follows from @Dave's comment that $A$ is a direct summand of $B$, as then $C\to\tau A$ is equal to the composition $C\to\tau A\to\tau B\to\tau A$, which is zero, since $C\to\tau A\to\tau B$ is zero, since it is the composition of two consecutive maps in a triangle. | |
| Aug 18 at 22:19 | history | edited | YCor | CC BY-SA 4.0 |
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| Aug 18 at 17:18 | comment | added | user267839 | @Dave Benson: On the contrary, I think that's the most "canonical" substitutor of the splitting lemma in triangulated world (...to seek in additive cats alone seems to be to weak as one could expect to have a nice criterion for splitting) one could expect, thanks! Maybe, one could subsequently ponder if the splitting in the example from derived cat actually motivated this question is a consequence of this criterion; this would require that the composition above $F \to f_{!}f^*F \to F$ fits actually in (or is homotopic to) a triangle whose 3th map is zero. Not sure if thats clear? | |
| Aug 18 at 16:54 | comment | added | Dave Benson | I'm not sure if this is what you're looking for, but one obvious statement is that if you have a map $A\to B$ in a triangulated category, complete it to a triangle $A \to B \to C \to \tau A$. Then $A\to B$ expresses $A$ as a direct summand of $B$ if and only if $C \to \tau A$ is the zero map. In this case we have $B\cong A\oplus C$. | |
| Aug 18 at 15:52 | history | edited | user267839 | CC BY-SA 4.0 |
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| Aug 18 at 15:41 | history | edited | user267839 | CC BY-SA 4.0 |
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| Aug 18 at 15:19 | history | edited | user267839 | CC BY-SA 4.0 |
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| Aug 18 at 14:43 | history | edited | user267839 | CC BY-SA 4.0 |
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| Aug 18 at 14:15 | history | edited | user267839 | CC BY-SA 4.0 |
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| Aug 18 at 14:07 | history | edited | user267839 | CC BY-SA 4.0 |
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| Aug 18 at 14:00 | history | asked | user267839 | CC BY-SA 4.0 |