Let $X$ be regular variety then it is known that $Q(Vect(X))\cong Q(\mathcal{M}_X)$. Where $Q$ is the Quillen's q-construction and $\mathcal{M}_X$ is the category of coherent sheaves on $X$. You can identify $Q(Vect(X))$ with a sub-complex of $Q(\mathcal{M}_X)$. Since this sub-complex is homotopy equivalent to the whole complex then it is a deformation retract. Is it possible to say where do the vertices (i.e. coherent sheaves) go under this deformation retract? Is it true that rank gets preserved? (for example the rank zero coherent sheaves go to the zero vector bundle.)(Rank of the coherent sheaf is defined after pulling it back to the generic point.)
More explanation: Since $Q(Vect(X))$ and $Q(\mathcal{M}_X)$ are $CW$ complexes and the zero cells of each are the vector bundles in one and coherent sheaves in the other. One is a tract of the other so by CW-approximation you can assume this retract maps vertices of $Q(\mathcal{M}_X)$ to the vertices of $Q(\mathcal{Vect}_X)$. There might not be a unique way of retracting but I think there is a canonical way of constructing the retract which depends on the proof of the resolution theorem and Quillen's theorem A. For simplicity assume that we take those coherent sheaves that the length of the resolution is 2 (You can find the proof for length 2 case in the Weibel's book it only uses Quillen theorem A twice) Quillen's theorem A is saying that if a functor $F: \mathcal{C}\rightarrow \mathcal{D}$ has this property that for all $d\in \mathcal{D}$ the $\mathcal{C} /d$ is contractible then $F$ induces homotopy equivalence. In order to construct an explicit retract I was thinking maybe it is possible by somehow gluing these data. If so then if $d$ is zero dimensional every thing that has a morphism to it is also zero dimensional so it has to retract to 0 (because of the q-construction you cannot map a higher rank vecrtex to a lower rank one.)
