5

I'm writing some notes on homological algebra and there I proved the long exact sequence in cohomology using the snake lemma. (I can give more details if anyone wants.) The next natural step is to prove that the connecting morphisms arise "functorially". In most books (Iversen, both books by Gelfand and Manin, both books by Kashiwara and Shapira, etc...) this is said to be either "obvious" or is an exercise to the reader.

Using elements (as Weibel does it), this is not too difficult but, since my main example of abelian category is the category of sheaves, I would rather avoid it.

Another possibility is to use the mapping cone to describe "explicitly" the connecting morphism and then the functoriality comes for free.

Because of this, I am thinking about using this (the need for an explicit description of the connecting morphisms) as a first motivation for the mapping cone construction. However, this would be very silly if the functoriality was indeed obvious. That's why I'm here: there is an "obvious" proof of this without using elements nor mapping cones?

(Not really related, but I would also love to know if someone can think of a better motivation for mapping cones, which I'll need to define the triangulated structure of the homotopic and derived categories.)

Gabriel
  • 4,513
  • 2
    Perhaps I’m missing something, but can’t you prove a “functorial snake lemma” by making sure everything is functorial, and then propagate the proof? (There’s even perhaps a sneakier shortcut, but I’m not sure it works – if $A$ is an abelian category, is the category $A’$ of $A$-morphisms $a \rightarrow a’$ abelian as well? It’s the category of $A$-valued functors on some (easy) category, so perhaps it’s true. If so, you can just consider complexes of objects of $A’$, which canonically are morphisms of complexes of objects of $A$, and have a LES in this category, ie a functorial LES in $A$). – Aphelli Feb 26 '21 at 12:38
  • The main motivation for mapping cones is making sure the derived category is triangulated. When you first study it, it will seem like a rather ad-hoc and pointless construction, but once you have distinguished triangles this is a rather natural construction that gives a lot of structure to your category. For example, functors of triangulated categories are somewhat exact, by preserving these triangles. – Sofía Marlasca Aparicio Feb 26 '21 at 12:47
  • @Mindlack This indeed seems to work, I'll try to do it! (And then come back here to share my findings, of course!) – Gabriel Feb 26 '21 at 13:19
  • 1
    @marlasca23 indeed triangles are very natural. But that they must be defined by mapping cones maybe isn't much more natural than "trying to describe explicitly the connecting morphisms", don't you think? – Gabriel Feb 26 '21 at 13:21

1 Answers1

3

This is not really an answer to my question but rather my understanding of the idea given by @Mindlack. (Proposition 5.4.2 is the fact that a functor category is abelian if the target is and "the previous theorem" is the existence of the long exact sequence in cohomology.)

I'm still very much interested in understanding better the motivation for the mapping cone.

(I've edited the answer to add the precise statement of the theorem.)

Gabriel
  • 4,513
  • 2
    Maybe I am picky and I sure am tired, but why does this imply functoriality? Supposed the snake morphism is not unique (I am unsure about this, maybe uniqueness is given and I am just blind), then this construction says that given a morphism of SES we find snakes compatible with the given morphism of complexes. But why is that functorial in the sense of being compatible with composition? – Jonas Linssen Feb 26 '21 at 17:38
  • When we say that the connecting morphisms are functorial we mean precisely that they define a functor from the category of short exact sequences of complexes to the category of complexes. This is precisely the fact that "given a morphism of SES we find snakes compatible with the given morphism of complexes", isn't it? (Also check the edit.) – Gabriel Feb 26 '21 at 17:44
  • 2
    My problem is that we have to make sure that we have a choice of connecting homomorphism for every SES. Then having those morphisms be compatible with morphisms of SES will indeed give functoriality by pasting squares. But as it stands the corollary only yields a connecting homomorphism for any particular given morphism of SES, so given a commuting triangle of SES, we might have different „functorial“ connecting homomorphisms for every side of the triangle... In other words, functoriality is not only about having assignments on Morphisms, but having them compatible with composition as well.. – Jonas Linssen Feb 26 '21 at 17:55
  • @PrudiiArca you are absolutely right. Do you know if the connecting morphisms are indeed functorial? I only ever saw what I wrote in the corollary in books. (And, for example, in the definition of delta-functor.)

    Also, is it true that the connecting morphism in the snake lemma is unique such that the bigger sequence is exact?

     – Gabriel Feb 26 '21 at 18:21
  • 1
    Im afraid I don’t. You are absolutely right with the fact that in approximately every textbook this tiny detail is ommitted. My gut says it ought to be true, backed by remembering that the snake lemma can be proven using pullbacks and pushouts (which I believe to be in lecture notes by Goedecke, I will look this up). Anyway your proof shows that for all practical purposes we can make a consistent choice for any diagram of SES, so even if it doesn’t define a functor per se, we can work with it as if it was :) – Jonas Linssen Feb 26 '21 at 18:31
  • I recall both P. Aluffi (in Algebra: Chapter 0) and S. Lang (in Algebra) saying that "the connecting morphisms define a functor from the category of short exact sequences of complexes to the category of complexes". But neither did prove it. Frankly, even "my corollary" was only proved (using elements) in Weibel's book, of all the (more than 10) references I've searched. – Gabriel Feb 26 '21 at 18:37
  • @PrudiiArca since this was distancing from the original purpose of this question, I've made a new one. https://math.stackexchange.com/questions/4041214/uniqueness-of-the-connecting-morphism-in-the-snake-lemma – Gabriel Feb 26 '21 at 19:53
  • I don’t know the general proof well, only the “module” proof with elements (which can indeed be done even with sheaves on sites) – but isn’t $\delta$ a canonical construction anyway, a definition that happens to have the satisfying LES property, like in Lemma 12.5.17 in https://stacks.math.columbia.edu/tag/00ZX ? – Aphelli Feb 26 '21 at 21:37
  • @Mindlack it is the unique map which satisfies some universal property (the one you cited) but it isn't the unique map which makes the long sequence exact. – Gabriel Feb 27 '21 at 09:24