Let $X$ be a manifold, $i: Z\to X$ is a closed embedding.

For a sheaf $S$ (of abelian groups) on a manifold $X$ and each $\varepsilon>0$ we denote by $Z_\varepsilon$ the set of points of $X$ that lie at distance $<\varepsilon$ from $X$. Consider the sheaf $S_\varepsilon:U\mapsto S(U\cap Z_\varepsilon)$, and also the section-wise limit $S_Z=\injlim_{\varepsilon\to 0} S_\varepsilon$.

I would like to understand the functor $W: S\to S_Z$ (I do not want to describe it section-wisely). Unfortunately, $S_Z$ is only an in-sheaf; it does not have to be a sheaf (in the 'usual' sense; note that the topological space $X$ is not noetherian). In particular, the stalk of $S_Z$ at any point $x\notin Z$ is easily seen to be $0$ (since $S_Z(U)=0$ for any $U$ such that its closure is disjoint from $Z$). On the other hand, the section $S_Z(X\setminus Z)$ does not seem to be $0$ (for example, if $S$ is constant); note that $X\setminus Z$ is not compact! See here the answer of algori to a previous version of this question (yesterday I believed that one can call $S_Z$ a sheaf).

Any hints (or references) for dealing with my $W$ would be very welcome! Note that $W$ is the limit of $i_{\varepsilon\ast}i_{\varepsilon}^\ast$. Unfortunately, the adjunctions that I wrote about previously probably do not hold (in any sense).

P.S. Some observations that do not seem to help me.

  1. My definition of $S_Z$ extends (without any changes) to presheaves.

  2. $W':P\mapsto P_Z$ sends ind-sheaves to ind-sheaves.

  3. It seems that ind-sheaves are (exactly) sheaves for the Grothendieck topology that admits only 'finite' covers.

So, perhaps one should use (somehow) the interplay between sheaves, ind-sheaves, and presheaves (and the corresponding topologies).

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    $\begingroup$ Mikhail -- I'm not sure I understand your construction of $S_Z$. It looks to me that if $\epsilon_1<\epsilon 2$, then $S_{\epsilon_1}$ maps to $S_{\epsilon_2}$, not the other way around, so the limit should be projective, not injective. In general, I don't know if $i_*i^{-1}$ has a left adjoint, but it has a right adjoint, `$i_!i^!$. $\endgroup$ – algori Oct 12 '11 at 19:29
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    $\begingroup$ Mikhail -- I don't see why $S_Z$ does not vanish on $X\setminus Z$. If say $dist(x,Z)=a>0$, then the stalk of any $S_\epsilon, \epsilon<a$ at $x$ is 0, and hence so is the stalk of the limit. I think that $S_Z$ is just $i_* i^{-1}S$. $\endgroup$ – algori Oct 12 '11 at 21:11
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    $\begingroup$ Dear Mikhail, if $i_\ast i^\ast$ has a left adjoint say $W$, then $i^\ast$ has a left adjoint as well, as one can see, using the fact that $i_\ast$ is fully faithful, as follows: we have $Hom(F,i^\ast G)=Hom(i_\ast F, i_\ast i^\ast G)=Hom(W i_\ast F, G)$. In other words, $W i_\ast$ is then a left adjoint of $i^\ast$. Therefore, the obstructions against the existence of $W$ are the same as the ones against the existence of a left adjoint of $i^\ast$. $\endgroup$ – Denis-Charles Cisinski Oct 12 '11 at 21:46
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    $\begingroup$ If you exclude infinite covers, you are entering unknown territory and should be careful. It's not even obvious what that means. I think the three most likely possibilities are: that they come for free; that you don't get a topos; or that you end up with sheaves on a weird space. $\endgroup$ – Ben Wieland Oct 12 '11 at 22:29
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    $\begingroup$ Dear Mikhail -- now I'm confused by the statement that $S_Z$ does not have to be a sheaf. By definition, if $A$ is any directed set and $F=(\{F_a\}_{a\in A}, \{f_a^b:F_a\to F_b\mid a,b\in A, a<b\}$ is a directed system of sheaves on some topological space, then the inductive limit $\mathop{\mathrm{inj}}\lim_{a\in A} F_a$ is the sheaf generated by the presheaf $U\to \mathop{\mathrm{inj}}\lim_{a\in A} F_a(U)$. This sheaf has all the expected properties (e.g. it is the colimit of $F$ in the category of sheaves; its stalks are the inductive limits of the stalks of $F_a$'s etc.), apart from one.. $\endgroup$ – algori Oct 13 '11 at 10:34

Let me show that $i^{-1}$ can't have a left adjoint when $X$ is a connected topological space and $Z\neq X$ is a point. From the remark by Denis-Charles Cisinski it would follow that $i_* i^{-1}$ can't have a left adjoint either.

Suppose $Z=\{x\}$ and $i^{-1}$ had a left adjoint $J$. Then we would have $$Hom (JF,G)=Hom (F,i^{-1} G)$$ for any $F$ a sheaf on $Z$ and $G$ a sheaf on $X$. Take a non-zero sheaf $F\in Sh(Z)$, i.e. a non-zero abelian group. Note that the stalk $(JF)_x\neq 0$ (this can be seen by taking $G$ to be the constant sheaf with stalk $F$). Let us show that for any $F$ the sheaf $JF$ must be supported at $x$. Suppose there is a $y\neq x$ such that $(JF)_y\neq 0$. Then take $G=i'_* i'^{-1}JF$ where $i'$ is the inclusion $\{y\}\to X$. We have $i^{-1} G=0$, and so the right hand side of the above adjunction formula is zero. The left hand side part is non-zero since the canonical map $JF\to i'_* i'^{-1}JF$ is non-zero.

So $JF$ must be supported on $x$. Take $G$ to be the constant sheaf with stalk $F$. This time ii is the left hand side of the formula that is zero (here we use that $X$ is connected and $Z\neq X$, so $JF=i_*i^{-1} JF$ can't map into the constant sheaf $G=\underline{F}_X$ in a non-zero way) and the right hand side that isn't.


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