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On page 362 of Ravi Vakil's notes, the author says

"It turns out that the main obstruction to vector bundles to be an abelian category is the failure of cokernels of maps of locally free sheaves - as $\mathcal O_X$-modules - to be locally free; we could define quasi-coherent sheaves to be those $\mathcal O_X$ modules that are locally cokernels..."

Indeed, the Stacks Project takes this as definition 10.1.

  1. Perhaps this is nitpicking, but I want to make sure - is this cokernel deficiency the only obstruction that needs fixing? What are some instructive examples displaying the failure of cokernels to be locally free?

  2. Since $\mathsf{QCoh}(X)$ is esentially defined as the minimal solution to the above obstruction, should it be viewed as the "universal solution" to removing it? In the case of finite rank locally free sheaves, $\mathsf{Coh}(X)$ is an even smaller abelian category containing them, so am I to understand quasicoherence is there to take care of all locally free sheaves?

Last nitpick - does definition 10.1 say $\mathcal F|_U$ is merely isomorphic to the cokernel object or that coupled with an epi it is an actual realization of the cokernel?

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  • $\begingroup$ The categorical stuff is not my bag, but for an example take $X=\mathbb A^1$, $\mathcal I$ the ideal of a point. The cokernel of $\mathcal I \to X$ is supported only on that point. $\endgroup$
    – John Brevik
    Aug 17, 2015 at 13:41
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    $\begingroup$ Because every quasicoherent sheaf is the cokernel of a map between locally free sheaves, it follows that every non-locally-free quasicoherent sheaf is an instructive example displaying the failure of cokernels to be locally free. More specifically, in the affine case, any non-projective module provides an example. $\endgroup$
    – WillO
    Aug 17, 2015 at 14:17
  • $\begingroup$ "every quasicoherent sheaf is the cokernel of a map between locally free sheaves" is not correct (unless for schemes with the resolution property). This is only correct locally. $\endgroup$
    – Martin Brandenburg
    Nov 3, 2015 at 7:00
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    $\begingroup$ Cokernels of morphisms between locally free sheaves are almost never locally free, for much the same reason that cokernels of linear maps between free modules are free. For instance the cokernel of the map $\mathbb{Z} \xrightarrow{\cdot 2} \mathbb{Z}$ is $\mathbb{Z}/(2)$, which is not free as a $\mathbb{Z}$-module. (In fact, the phrase "for much the same reason" can be made entirely rigorous, by something called "the internal language of a topos".) $\endgroup$
    – Ingo Blechschmidt
    Nov 29, 2015 at 21:47
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    $\begingroup$ This might help: mathoverflow.net/questions/150168/… $\endgroup$
    – Saal Hardali
    Apr 24, 2016 at 8:38

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