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Jun 7, 2018 at 16:04 answer added Christophe Wacheux timeline score: 2
Jun 6, 2018 at 17:06 history edited Christophe Wacheux CC BY-SA 4.0
Canceling my previous edition: I was right from the beginning !
Jun 6, 2018 at 16:10 history edited Christophe Wacheux CC BY-SA 4.0
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Jun 6, 2018 at 16:07 comment added Christophe Wacheux Actually, it is not true that $\Gamma(X,\mathbb{C}_\mathcal{D}) = \mathcal{D}$. Sorry, my mistake. I will edit.
Jun 6, 2018 at 15:56 history edited Christophe Wacheux CC BY-SA 4.0
initial topology is the coarsest, not the thinnest topology making the function continuous
Jun 5, 2018 at 7:47 comment added user40276 This probably works as a counter-example to flabiness and softness. It's not an answer. It's a sketch of an idea that I'm too lazy to work out all the details and it may or may not work. About $X$ being a circle, the idea would work for any closed path in an arbitrary $X$ covered by more than $2$ open sets. The idea is that you can start extending clockwise (or counter-clockwise), but, when you reach the end of the path, it doesn't agree with its initial value.
Jun 4, 2018 at 8:44 comment added Christophe Wacheux @user40276: I am not exactly sure of which question you are giving an answer ? Is this some counter-example ? To what softness, flabiness etc. ? Also, is it clear I can find a smooth algebra with $f$ and $g$ as you mention ? Also, I don't get how the fact that $X$ is a circle should intervene.
Jun 1, 2018 at 14:55 history edited Christophe Wacheux CC BY-SA 4.0
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Jun 1, 2018 at 14:30 history edited Christophe Wacheux CC BY-SA 4.0
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May 31, 2018 at 23:38 comment added user40276 To get an example with other kind of topologies just keep adding open sets by adding other smooth functions on the circle and also choose $f$ and $g$ very non-smooth so that no function agrees with $f$ and $g$ outside $U \cup V$.
May 31, 2018 at 23:29 comment added user40276 I'm too lazy to write down everything in the general case and I also don't have a scanner now. So let me just sketch the picture that I have in mind: pick $X = S^1$ and $3$ connected open sets $U, V$ and $W$, each one intersecting exactly two open sets. Think about these sets as open sets of $X$. Pick two functions $f$ and $g$ that agree only on $U \cap V$ and no other function agrees with $f$ or $g$ outside of $U \cup V$. Define a new function $h$ on $U \cup V$ that agrees with $f$ on $U$ and $g$ on $V$, then you can't extend $h$ (you get a kind of holonomy).
May 31, 2018 at 23:17 comment added user40276 Ok. I can understand most of your notation after your edit. However that $P_X$ on the definition of the sheaf should probably be $P_U$, otherwise your question is trivial. In any case, I'm almost sure that none of the properties that you want will be satisfied even when you restrict the conditions on the topology of $X$. I'm less sure about the acyclicity under some conditions.
May 31, 2018 at 7:38 history edited Christophe Wacheux CC BY-SA 4.0
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May 31, 2018 at 7:21 comment added Christophe Wacheux Given the usual topology on $\mathbb{R}$, the topology $\tau_\mathcal{D}$ is defined as the initial topology with respect to $D$, i.e. the thinnest topology on $X$ that makes all maps in $\mathcal{D}$ continuous, so it is a topology on $X$. As for your second comment, you're right, but that's why I mentionned the answer might require extra topological assumptions: if there is a well-known typology for $\mathbb{C}_\mathcal{D}$, but only if $(X,\tau_\mathcal{D})$ is paracompact, then it is ok for me.
May 30, 2018 at 1:12 comment added user40276 What's the topology on $X$? $x \in U \in \tau_D$, but $\tau_D$ was defined as a topology on $D$ and not on $X$. In any case, the property of being fine is only defined, as far as I know, for paracompact Hausdorff spaces.
May 29, 2018 at 15:21 history asked Christophe Wacheux CC BY-SA 4.0