TITLE: Necessity of hypercovers for sheaf condition for simplicial sheaves
QUESTION [20 upvotes]: I'm trying to understand where the definition of simplicial sheaf on a space/site comes from.  
For a presheaf $F$ of sets on a topological space $X$, the sheaf condition can be viewed as saying that $F$ is a 'local object' with respect to the maps of presheaves $colim(\coprod U_{ij} \underrightarrow{\rightarrow} \coprod U_{i}) \rightarrow X$'for every open cover $\{ U_{i} \rightarrow X \}$. That is to say, HOMing $colim(\coprod U_{ij} \underrightarrow{\rightarrow} \coprod U_{i}) \rightarrow X$' into $F$ gives an isomorphism, and that is easily seen to be the usual sheaf condition. (This seems to be a fairly well-known way to say the sheaf condition, and I learned it from the nlab.)
Now for $F$ a presheaf of simplicial sets, people require more for the sheaf condition, namely that $F$ is local with respect to $hocolim(U_{*}) \rightarrow X$', where $U_{*}$ is a so-called hypercover, and these are some sort of refinement of simplicial diagrams that can be built from multiple intersections of elements of an open cover $U_{i} \rightarrow X$. (I don't really know how to draw a simplicial diagram here. Sorry.) 
My question is why in the definition of simplicial sheaf it is necessary to consider not just double intersections of the $U_{i}$, but all intersections (organized into a simplicial diagram). I've read people say, 'The condition is satisfied for spaces, so should be satisfied for general sheaves', but I don't find this a completely convincing reason, since we don't need (?) the whole simplicial diagram to reconstruct a space (at least if we take naive colimits instead of homotopy colimits).
I imagine good answers would be something like 'it is necessary for homotopy invariance of constructions' or 'here is a presheaf F that we'd like to be a sheaf, but HOMing only double intersections into it doesn't see enough information'.

REPLY [6 votes]: This answer is not really different from David's or Marc's; I just want to boil things down.
In brief: 
If you want "weak equivalence of simplicial sheaves" to mean the same thing as "induce weak equivalence on stalks", you must localize simplicial presheaves with respect to hypercovers.  
You could only localize with respect to covers, but then it may be possible for there to be an object whose stalks are all weakly contractible, but which is not itself weakly equivalent to the terminal object.
Note: a "weak equivalence on stalks" amounts to the same thing as "isomorphism of homotopy-group sheaves", so I could have instead said: if you only localize with respect to covers, then it may be possible for there to be an object with trivial homotopy-group sheaves, but which is itself not weakly equivalent to the terminal object.
I highly recommend the paper of Dugger, Hollander, and Isaksen, Hypercovers and simplicial presheaves, which explains this, and gives an example exhibiting the phenomenon I just mentioned.  
Added.
For "localize with respect to covers", I mean: from an open cover of an open set $X$ build a simplicial object $C_\bullet$ of presheaves, with $C_0=\coprod U_i$, $C_1=\coprod U_i\cap U_j$, etc.  Then a simplicial presheaf $F$ has descent for covers if 
$$F(X) \to \mathrm{holim}  F(C_\bullet)$$
is a weak equivalence of simplicial sets (for all open covers $\{U_i\}$ of open sets $X$). Note that I'm taking a homotopy limit of a cosimplicial diagram here.
If you just use the "first two steps of the diagram", you would be saying that the map $F(X)$ to the "homotopy equalizer" of $F(C_0) \rightrightarrows F(C_1)$ is a weak equivalence.  This is a different condition, and is the wrong way.  
Here is one way to see why it is wrong (I'll be a bit imprecise here): start with the constant presheaf whose value is an Eilenberg-MacLane space $K(G,n)$, and "localize" it to get a simplicial presheaf $F$ which has descent for covers.  I want to be able to say that $F(X)$ has something to do with the cohomology of $X$.  With the correct definition (full simplicial diagram), you can see that you are seeing something that looks similar to the classical definition of Cech cohomology.  In fact, if $\{U_i\}$ is a "good cover", so that all finite intersections are either empty or contractible, this is  what you get.  If you just use the "two-step" definition, it's not clear that $F(X)$ will know anything about 3-fold intersections of open sets in the cover, so presumably can't compute Cech cohomology.