Timeline for Reducing the simplices in the nerve of a category with an object with trivial endomorphism monoid
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 30, 2014 at 14:46 | vote | accept | Werner Thumann | ||
| Mar 25, 2014 at 9:09 | comment | added | Werner Thumann | Could you give me a reference for this? | |
| Mar 25, 2014 at 5:45 | comment | added | მამუკა ჯიბლაძე | ...and then I believe it is actually a particular case of splitting an idempotent ($A\to X\to A$ in your case), which is known not to affect homotopy type... | |
| Mar 25, 2014 at 5:42 | comment | added | მამუკა ჯიბლაძე | Oh I see, sorry, you took care of that - if $N^r(C)$ differs from $N(C)$ then there actually $X\to A\to X$ must occur with $A\ne X$... | |
| Mar 25, 2014 at 5:25 | comment | added | მამუკა ჯიბლაძე | Whoops wait! @ViditNanda has a serious point here (although the given example did not work): consider the case when $C$ is the cone over $C^-$, i. e. there is a single morphism from $X$ to any other object and no morphisms from other objects to $X$. Then $C$ is contractible while $C^-$ can be arbitrary! | |
| Mar 25, 2014 at 5:20 | comment | added | მამუკა ჯიბლაძე | Actually strictly speaking you need also to consider separately the case when $X$ forms a separate connected component, since in this case homotopy types of $C^-$ and $C$ are different. | |
| Mar 24, 2014 at 22:22 | comment | added | Werner Thumann | As pointed out by მამუკაჯიბლაძე, the arrow $b\rightarrow f$ closes the circle in the full subcategory where $a$ is not present, it is also a $S^1$. | |
| Mar 24, 2014 at 19:32 | vote | accept | Werner Thumann | ||
| Mar 30, 2014 at 14:46 | |||||
| Mar 24, 2014 at 19:29 | vote | accept | Werner Thumann | ||
| Mar 24, 2014 at 19:29 | |||||
| Mar 24, 2014 at 19:25 | comment | added | მამუკა ჯიბლაძე | @ViditNanda Both identifying isomorphic objects and leaving less (but at least one) in each isomorphism class gives categories equivalent to the original one | |
| Mar 24, 2014 at 19:19 | comment | added | Vidit Nanda | @მამუკაჯიბლაძე Agreed, but that doesn't change the homotopy type of the classifying space. Essentially, you can contract $b$ to $a$ via the obvious natural transformation and update all the morphisms accordingly -- this is not quite the same as deleting $a$ from the category altogether! | |
| Mar 24, 2014 at 18:58 | comment | added | მამუკა ჯიბლაძე | @ViditNanda These cannot be the only morphisms - e. g. there must be composite $b\to a\to f$ | |
| Mar 24, 2014 at 18:35 | comment | added | Vidit Nanda | Depending on what "kick the $X$" means, the final upshot in your answer does not appear to be true. Consider a category with $6$ objects $a \leftrightarrow b \rightarrow c \leftarrow \cdots \rightarrow f \leftarrow a$ where the maps between $a$ and $b$ are invertible (hence bidirectional) but the other maps are not. The classifying space with $a$ present is homotopy-equivalent to $S^1$ but if we remove $a$ altogether we get a contractible category. | |
| Mar 24, 2014 at 18:03 | history | edited | Werner Thumann | CC BY-SA 3.0 |
added 379 characters in body
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| Mar 24, 2014 at 17:44 | history | answered | Werner Thumann | CC BY-SA 3.0 |