Skip to main content
MathOverflow

Return to Answer

Added comment on relation to the other answer.
Source Link
Torsten Ekedahl
Torsten Ekedahl
  • 22.6k
  • 2
  • 81
  • 98

[Added] Some comments intended to give some kind of relation with the answer provided by fpqc. My suggested answer is not homotopy invariant in the sense that a weak (or even homotopy) equivalence of simplicial sets does not induce an equivalence on the category of sheaves. This is so however if one, as per above, adds the condition that all the maps $T_{f(c)} \to T_c$ are isomorphisms. However, that condition is not so good as many maps that are not weak equivalences induces category equivalences (it is enough that the map induce isomorphisms on $\pi_0$ and $\pi_1$). This is a well-known phenomenon and has to do with the fact the $T_c$ are just sets. One could go further and assume that the $T_c$ are topological spaces and the maps $T_{f(c)} \to T_c$ continuous. Of course adding the condition that these maps be homeomorphisms shouldn't be right thing to do, instead one should demand that they be homotopy (or weak) equivalences. Again, this shouldn't be quite it because of the transitivity conditions. We should not have that the composite $T_{g(f(c))} \to T_{f(c)} \to T_c$ should be equal to $T_{g(f(c))} \to T_c$ but rather homotopic to it. Once we have opened that can of worms we should impose higher homotopies between repeated composites. This can no doubt be (has been) done but there seems to be an easier way out. In the first step away from set-valued $T_c$ we have the possibility of they being instead categories. In that case the higher homotopy conditions is that we should have a pseudofunctor $\Delta/F \to \mathcal{C}\mathrm{at}$. Even they are somewhat unpleasant and it is much better to pass to the associated fibred category $\mathcal{T} \to \Delta/F$. In the general case, and admitting that $\Delta/F$ is essentially the same things as $F$ itself, we should therefore look at (Serre) fibrations $X \to |F|$ or if we want to stay completely simplicial, Kan fibrations $X \to F$. This gives another notion of (very flabby) sheaf which now should be homotopy invariant (though that should probably be in the sense of homotopy equivalence of $\Delta$-enriched categories).

[Added] Some comments intended to give some kind of relation with the answer provided by fpqc. My suggested answer is not homotopy invariant in the sense that a weak (or even homotopy) equivalence of simplicial sets does not induce an equivalence on the category of sheaves. This is so however if one, as per above, adds the condition that all the maps $T_{f(c)} \to T_c$ are isomorphisms. However, that condition is not so good as many maps that are not weak equivalences induces category equivalences (it is enough that the map induce isomorphisms on $\pi_0$ and $\pi_1$). This is a well-known phenomenon and has to do with the fact the $T_c$ are just sets. One could go further and assume that the $T_c$ are topological spaces and the maps $T_{f(c)} \to T_c$ continuous. Of course adding the condition that these maps be homeomorphisms shouldn't be right thing to do, instead one should demand that they be homotopy (or weak) equivalences. Again, this shouldn't be quite it because of the transitivity conditions. We should not have that the composite $T_{g(f(c))} \to T_{f(c)} \to T_c$ should be equal to $T_{g(f(c))} \to T_c$ but rather homotopic to it. Once we have opened that can of worms we should impose higher homotopies between repeated composites. This can no doubt be (has been) done but there seems to be an easier way out. In the first step away from set-valued $T_c$ we have the possibility of they being instead categories. In that case the higher homotopy conditions is that we should have a pseudofunctor $\Delta/F \to \mathcal{C}\mathrm{at}$. Even they are somewhat unpleasant and it is much better to pass to the associated fibred category $\mathcal{T} \to \Delta/F$. In the general case, and admitting that $\Delta/F$ is essentially the same things as $F$ itself, we should therefore look at (Serre) fibrations $X \to |F|$ or if we want to stay completely simplicial, Kan fibrations $X \to F$. This gives another notion of (very flabby) sheaf which now should be homotopy invariant (though that should probably be in the sense of homotopy equivalence of $\Delta$-enriched categories).

Source Link
Torsten Ekedahl
Torsten Ekedahl
  • 22.6k
  • 2
  • 81
  • 98

Clearly looking at sheaves on the geometric realisation gives something too far removed from the simplicial picture. This is essentially because there are too many sheaves on a simplex have (most of which are unrelated to simplicial ideas). What one could do is to consider such sheaves which are constructible with respect to the skeleton filtration, i.e., are constant on each open simplex. This can be described inductively using Artin gluing. I think it amounts to the following for a simplicial set $F$.

For each simplex $c\in F_n$ we have a set $T_c$, the constant value of the sheaf $T$ on the interior of the simplex corresponding to $c$.

For each surjective map $f\colon [n] \to [m]$ in $\Delta$ the corresponding (degeneracy) map on geometric simplices maps the interior of $\Delta_n$ into (onto in fact) the interior of $\Delta_m$ and hence we have a bijection $T_{f(c)} \to T_c$. These bijections are transitive with respect to compositions of $f$'s.

For each injective map $f\colon [m] \to [n]$ in $\Delta$ the corresponding map on geometric simplices maps $\Delta_m$ onto a closed subset of $\Delta_n$. If $j\colon \Delta^o_n \hookrightarrow \Delta_n$ is the inclusion of the interior we get an adjunction map $T \to j_\ast j^\ast T$ and $j_\ast j^\ast T=T_c$ where $T_c$ also denotes the constanct sheaf with value $T_c$. If $f'\colon \Delta^o_m \hookrightarrow \Delta_n$ is the inclusion of the interior composed with $f$ we can restrict the adjunction map to get a map $T_{f(c)}=f'^\ast T \to f'T_c$ and taking global sections we get an actual map $T_{f(c)} \to T_c$. These maps are transitive with respect compositions of $f$'s.

We have a compatibility between maps coming from surjections and injections. Unless something very funny is going on this compatibility should be that we wind up with a function on the comma category $\Delta/F$ which takes surjections $[n] \to [m]$ to isomorphisms.

There is the stronger condition on the sheaf $F$, namely that it is constant on each star of each simplex. This means on the one hand that it is locally constant on the geometric realisation, on the other hand that $T_{f(c)} \to T_c$ is always an isomorpism.