Let $X$ be a manifold (or a variety; then I am interested in etale sheaves), $i: Z\to X$ is a closed embedding. I need a left adjoint to the functor $i_\ast i^\ast$ (for sheaves of abelian groups on $X$).

It seems that such an adjoint was not considered 'classically'. Yet now I will explain why it exists (at least, in the 'topological' setting).

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 limit sheaf $S_Z=\injlim_{\varepsilon\to 0} S_\varepsilon$. Then for any sheaf $F/X$ we have $Hom(S_Z,F)\cong Hom(i^\ast S,i^\ast F)$, i.e. the functor $W:S\mapsto S_Z$ is the adjoint desired.

My question is: does there exist a 'sheaf-theoretic' description of $W$ (without describing each section of $S_Z$ separately, and without mentioning $Z_\varepsilon$!)? Let $i_\varepsilon$ denote the (open) embedding $Z_\varepsilon\to X$. Then $W$ is the limit of $i_{\varepsilon\ast}i_{\varepsilon}^\ast$. Certainly, the problem would be solved if $i^\ast$ possessed a left adjoint; yet it probably does not. In particular, $W$ cannot factorize through $i_*$ since otherwise $S_Z$ would be zero on $X\setminus Z$ (and this is wrong already for a constant $S$).

Any hints (or references) for dealing with my $W$ would be very welcome! I wouldn't like to introduce any 'infinitesmall' topology here (yet comments in this direction could also be quite interesting).

Some cautions:

1. $i_{\varepsilon}$ is an open embedding, and not a closed one.

2. $Z_{\varepsilon}\cap U$ is usually larger than $(Z\cap U)_{\varepsilon}$.

3. When I speak about sheaves, I only allow coverings by manifolds (and not by infinite disjoint unions of those). $S_Z$ is (always) a sheaf only in this restricted sense (so, it is somewhat pathological). Note that $S_Z(U)=0$ for any $U$ such that its closure is disjoint from $Z$. This does not seem to imply that $S_Z(X\setminus Z)=0$; note that $X\setminus Z$ is not compact!