Let $A$ be a finite dimensional algebra . Let $P\_{\inf}$ be the full subcategory of modules having finite projective dimension and $P\_r$ the subcategory of modules having projective dimension bounded by $r$. Does $P\_{\inf}$ being contravariantly finite imply that $P\_r$ is contravariantly finite for all $r$?



Contravariant finiteness of subcategories
Not in general, although there may be some classes of algebras where it holds. For a simple counterexample $A$ consider the path algebra (over an infinite field) of the quiver $$1 \stackrel{a, b}\Rightarrow 2 \stackrel{c}{\rightarrow} 3$$ modulo the relation $ca=0$. This algebra has global dimension 2, so in particular the subcategory of modules of finite projective dimension is all of $A$-mod, which is contravariantly finite. However, the simple module $S\_1$ has no right-approximation by a module of projective dimension $\leq 1$, so $P\_1$ is not contravariantly finite. Informally, one sees there is a one-parameter family of nonisomorphic indecomposables with dimension vector (1,1,0), each of which (with one exception) has projective dimension 1, and maps onto $S\_1$. Then one can check that there is no single finite-dimensional module of projective dimension 1 through which all of these maps can factor.


I believe that one can make this argument more rigorous (but I haven't checked the details) by using the fact that this algebra $A$ is stably equivalent (by constructing a node) to the algebra $\Lambda$ studied by Igusa, Smalo and Todorov in their paper "Finite projectivity and contravariant finiteness," Proc. Amer. Math. Soc. 109 (1990), no. 4, 937-941.


