5
$\begingroup$

In an abelian category $\mathscr A$ we encounters the notions of kernel, cokernel, chain homology, derived functors, etc. These notions are frequently referred to as functors, and yes, they actually behave functorially.

But lately it comes to my mind that, for example, which object is $H_n(X)$ for $X_\bullet$ a chain complex in $\mathscr A$, exactly? Formally it is the cokernel $\ker\partial_n/\operatorname{im}\partial_{n+1}$; one may realize it with some kernel and cokernel "functors", whose target is the arrow category $\text{Arr}(\mathscr A)$, and applying the functor that takes codomain $\text{Arr}(\mathscr A)\to\mathscr A$ to finally obtain an object in $\mathscr A$. The last functor being unambiguous, it suffices to figure out the precise meaning of kernel and cokernel functors; by duality we focus on the kernel functor.

$\ker$ should be a functor $\text{Arr}(\mathscr A)\to\text{Arr}(\mathscr A)$(e.g. in Section 3 of this nLab page), but I've never seen it defined explicitly. For a morphism $f:A\to B$, which arrow is $\ker f$, among ones that are isomorphic to each other by a unique isomorphism? In $\text{Ab}$ or $\text{$R$-Mod}$ there is a canonical construction, but what about a general abelian category? If we just choose for each $f$ an arbitrary $\ker f$ among all isomorphic ones, wouldn't this involve an ultimately strong form of Axion of Choice?

The same issue arises when we consider the derived category. For example, assuming $\mathscr A$ has enough projectives, we claim that there is a canonical functor $F:\mathscr A\to\mathscr D_{\bullet}(\mathscr A)$, by sending any object $A$ to its projective resolution, which we argue to be well-defined by showing that projective resolution of $A$ is unique up to a unique isomorphism. But which chain is $F(A)$ exactly?

Question: is there a standard way in category theory to resolve this? Do we form a new category by identifying those objects that are isomorphic in some sense? I could not really formalize this idea; when I tried to work it out for $\ker$, it seems that domain functor is no longer well-defined(with target in $\mathscr A$) in the new category, so homology cannot be defined in this way. I feel completely lost.

Thanks for any advice or reference.

$\endgroup$
  • $\begingroup$ Categorists always seem to gloss over such issues. But identifying objects that are isomorphism is liable to cause more problems than it solves. $\endgroup$ – Lord Shark the Unknown Oct 13 '18 at 9:45
  • 2
    $\begingroup$ There are different ways of adressing this issue : one is using an "axiom of choice for classes"; but another one is considering $\ker$ for instance and projective resolution as given data in the category (maybe not "projective resolution" but "a projective that surjects onto $A$, with an epimorphism that witnesses that). In practice, the categories we use actually have such data (in $Ab$, the $\ker$ is precisely the subgroup, the coker is the quotient etc.) $\endgroup$ – Max Oct 13 '18 at 9:45
  • $\begingroup$ @Max I feel the role of axiom of choice really strange here: if the category is small, then the said "axiom of choice for classes" reduces to the ordinary axiom of choice (for sets), but in this case we may just apply Freyd-Mitchell embedding (the proof of which does not rely on axiom of choice as far as I can remember) and work with the canonical construction of $\ker,\operatorname{coker}$ in $\text{$R$-Mod}$. $\endgroup$ – Cave Johnson Oct 13 '18 at 9:57
  • 1
    $\begingroup$ Yes, for small categories nothing of that sort is necessary, but for large categories it may be useful. For you second question, yes that's what that would mean (or, better, an abelian category is blabla together with kernel, cokernel, ... functors). I don't have a reference but for instance Hovey treats model categories this way : instead of saying "there exists a factorization", the paper says "a model category comes with factorization functors" (I don't have a reference but I jave a feeling it's a classical enough method that it'd be easy to find references) $\endgroup$ – Max Oct 13 '18 at 10:14
  • 1
    $\begingroup$ One term for what Max is talking about is having "chosen products/limits/colimits/etc." which indeed means the category is equipped with a suitable functor. I'm pretty sure this is discussed in "Categories for the Working Mathematician". At least for the areas of category theory I haunt, I would say that this is usually implicitly what categorists mean when they say a category "has limits" or whatever, or at least it is taken for granted that such functors can be produced. $\endgroup$ – Derek Elkins Oct 13 '18 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.