Let $X$ be a nontrivial ringed space (i.e. all stalks are nonzero). To every locally free module $M$ on $X$ of constant rank $n$ we can associated it's determinant $\det(M)$, which is a line bundle and is defined as the $n$th exterior power of $M$. We can also define $\det(M)$ if the rank is not assumed to be constant. It's surely locally constant, so $X$ is partitioned into the open subsets $(X\_n)\_{n \geq 0}$, where $M$ has constant rank $n$, and we can glue the $\det(M|\_{X\_n})$ to get the determinant $\det(M)$. There is no doubt that this is well-defined, but it seems to me a bit uncanonical.


For example look at the functoriality: If $f : M \to N$ is a homomorphism, then $\det(f) : \det(M) \to \det(N)$ is first defined on the intersections on the open subsets on which $\det(M)$ and $\det(N)$ are defined, and then glued. Then a small argument is needed to prove that this is, indeed, a functor. Isn't this ugly? Therefore:


**Question** Is there a characterization of the functor $\det(-)$ from locally finite free modules to line bundles which does not depend on partitions of $X$?


For example, for every $n \geq 0$, the functor $\det(-)$ on locally free modules of rank $n$ has a universal property, namely $\text{Hom}(\det(M),L)$ corresponds to alternating maps $M^n \to L$. You can write down a similar universal property when $n$ is a fixed locally constant function on $X$. But my question is if we can do it without fixing $n$.



universal property of the determinant bundle
There is a paper by Knudsen and Mumford that gives a thorough treatment of determinants of perfect complexes. Mumford has put a copy online [here](http://www.dam.brown.edu/people/mumford/Papers/DigitizedAlgGeomPapers--ForNon-CommercialUse/76b--DetDiv-Knudsen.pdf).


**Edit:** [Knudsen's 2002 paper](http://projecteuclid.org/euclid.mmj/1028575741) that Theo Buehler linked in the comment below appears to resolve the question in a more canonical way than the older paper. It also has a neat letter by Grothendieck at the end.


