Re-bonjour, Thank you for your answers. My question was very general. So here is the example. I am going to define the category C and the collection of morphisms S, with respect to what I would like to localize. The object of C are the oriented graph. Such an object X is a topological space obtained by choosing a discrete set X^0 and by attaching 1-dimensional cells *with orientations*. It is a 1-dimensional CW-complex with oriented arrows. The morphisms of C are the continuous maps f from X to Y satisfying this conditions : 1) f(X^0)\subset Y^0 2) f is orientation-preserving 3) f is non-contracting in the sense that a 1-cell is never contracted to one point. Remark I : in C, an arrow x--> is not isomorphic to a point. Remark II : an arrow a-->b can be mapped on the loop a-->a with one oriented arrow from a to a. A morphism f of C is in S if and only if f induces an homeomorphism on the underlying topological spaces. Now here is an example of f\in S which is not invertible : a--->b mapped on a-->x-->b This morphism has no inverse in C because the image of x must be equal to a or b by 1) and therefore one of the arrows would be contracted by 2), which contredicts 3). I would like to know if C[S^{-1}] exists or no (in the same universe). The irresistible conjecture is of course that C[S^{-1}] is equivalent to the category whose objects are that of C and whose morphisms from A to B are the subset of C^0(A,B) (the set of continuous maps from A to B) containing all composites of the form g_1.f_1^{-1}.g_2...g_n.f_n^{-1}.g_{n+1} where g_1,...,g_{n+1}, are morphisms of C and f_1,...,f_n morphisms in S. The Ore condition is not satisfied by S because of this example. The Ore condition says that for any s:A-->B in S, and any f:X-->B, there exists t:Y-->X in S and g:Y-->A such that s.g=f.t. Now the counterexample : A is a--->b, B is a-->x-->b with s as above ; X is a-->x with the inclusion f from X in B. Then necessarily Y=X and t=Id. And s.g(x)=b and f.t(x)=x. Thanks in advance. pg.