Timeline for What does a partial map classifier look like as a sheaf?
Current License: CC BY-SA 4.0
5 events
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| Dec 8, 2019 at 15:47 | comment | added | Simon Henry | Alternatively, you can use internal logic to prove the following: Given $V \subset U$ and $V \rightarrow S$, there is a unique map $U \rightarrow S^{\perp}$ which makes the square a pullback. This shows that (externally) maps from an arbitrary object $U$ so $S^\perp$ are the same as pairs $V \subset U$ and $f:V \rightarrow U$. You then get the desired result by applying this to $U$ a representable sheaf. | |
| Dec 8, 2019 at 11:14 | comment | added | wlad | @მამუკაჯიბლაძე Thanks! | |
| Dec 8, 2019 at 11:13 | comment | added | მამუკა ჯიბლაძე | Conversely a pair $(V,f)$ as in your guess gives a sheaf map $[V]\to S$ which in turn determines an inclusion $(i,[f]):[V]\rightarrowtail[U]\times S$, where $i:[V]\to[U]$ corresponds to the inclusion of $V$ into $U$ and $[f]:[V]\to S$ corresponds to $f$. | |
| Dec 8, 2019 at 11:07 | comment | added | მამუკა ჯიბლაძე | Look at $S_\bot$ as a subsheaf of $\mathcal P(S)$. Sections of $\mathcal P(S)$ over an $U$ correspond to arbitrary subsheaves $R\subseteq[U]\times S$. Here $[U]$ is the sheaf corresponding to $U$, it has a single section over $V\subseteq U$ and no sections over $V\nsubseteq U$. Sections of $S_\bot$ over $U$ correspond to such $R$ that the composite $R\subseteq[U]\times S\to[U]$ with the projection is an inclusion. This inclusion then identifies $R$ with a subsheaf of $[U]$ which always has form $[V]$ for some $V\subseteq U$, and $R\subseteq[U]\times S\to S$ determines a section of $S$ over $V$ | |
| Dec 8, 2019 at 10:39 | history | asked | wlad | CC BY-SA 4.0 |