Homological Algebra Puzzle

December 2, 2010

Posted by John Baez

$MathML-enabled post (click for more details).$

James Dolan gave me another puzzle today.

This one is a bit more sophisticated. Can you find a really nice solution? I’m also curious to hear how well-known this fact is. (Neither of us know a reference.)

$MathML-enabled post (click for more details).$

Suppose $A$ is an abelian category and let $Arr(A)$ be the arrow category of $A$ . There are obvious functors $ker, coker : Arr(A) \to Arr(A)$ . Prove that $coker$ is the left adjoint of $ker$ .

Or if that’s too jargon-packed, let me expand it out a bit for you:

Suppose $A$ is an abelian category. Let $Arr(A)$ be the category where an object is a morphism $f : a \to b$ in $A$ , and a morphism is a commutative square in $A$ . There are functors

$ker, coker : Arr(A) \to Arr(A)$

where $ker$ sends any morphism $f : a \to b$ in $A$ to the obvious morphism

$ker(f) \to a$

while $coker$ sends it to the obvious morphism

$b \to coker(f)$

Show that the functor $coker$ is left adjoint to the functor $ker$ .

And here’s a question for the experts: if we have an additive category with kernels and cokernels, and $ker$ is right adjoint to $coker$ in the above sense, is it abelian? If so, is this characterization already known?

Posted at December 2, 2010 11:28 AM UTC

TrackBack URL for this Entry: https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2317

10 Comments & 0 Trackbacks

Re: Homological Algebra Puzzle

This adjunction holds in any category enriched in pointed sets with all kernels and cokernels (so it doesn’t imply abelianness). It’s a special case of a more general scheme: in general, quotients of some kind are left adjoint to the corresponding kernels.

This general theory was developed in “Factorizations in bicategories”, by Betti, Schumacher and Street, which was never published, but Betti has a paper in JPAA where it is described: “Adjointness in descent theory”. If I remember correctly, they attribute the general adjunction to Robert Paré. Their proof is, I think, very nice (it takes one line, once kernels and quotients are defined in the appropriate way). You can also look at Section 1.2.2 (and around) of my PhD thesis.

Posted by: Mathieu Dupont on December 2, 2010 1:43 PM | Permalink | Reply to this

Re: Homological Algebra Puzzle

$MathML-enabled post (click for more details).$

I think essentially the same general theory is described on the nLab at generalized kernel. This puzzle is even one of the examples mentioned!

Posted by: Mike Shulman on December 2, 2010 9:45 PM | Permalink | Reply to this

Re: Homological Algebra Puzzle

$MathML-enabled post (click for more details).$

my PhD thesis.

Do abelian categories in dimension 2 sit inside stable $\infty$ -categories in the way one would hope?

Posted by: Urs Schreiber on December 2, 2010 10:38 PM | Permalink | Reply to this

Re: Homological Algebra Puzzle

$MathML-enabled post (click for more details).$

The first “puzzle” is not hard; do I need to do the rot13 thing?

Given arrows $f: a \to b$ , $g: c \to d$ , one can show that maps from $coker(f)$ to $g$ are in natural bijective correspondence with maps $h: b \to c$ such that $g \circ h \circ f = 0$ . By dualizing that statement, one gets that these are in natural bijective correspondence with maps from $f$ to $ker(g)$ . Off-hand I do not know a reference.

The answer to the second question has to be “no”, because the solution to the puzzle has nothing to do with the fact that monomorphisms are kernels, etc. For an counterexample, try topological abelian groups, where not every mono is a kernel, as you can see from the irrational torus.

Posted by: Todd Trimble on December 2, 2010 1:44 PM | Permalink | Reply to this

Re: Homological Algebra Puzzle

$MathML-enabled post (click for more details).$

Er, I mean the irrational line on a torus, the map $\mathbb{R} \to \mathbb{R}/\mathbb{Z} \times \mathbb{R}/\mathbb{Z}$ defined by $x \mapsto (a x, b x)$ where $a/b$ is irrational.

Posted by: Todd Trimble on December 2, 2010 1:48 PM | Permalink | Reply to this

Re: Homological Algebra Puzzle

$MathML-enabled post (click for more details).$

do I need to do the rot13 thing?

No. Let’s all stop investing energy into hiding information and instead invest it into making information available.

If John keeps posting little facts and people who like to solve them write out the proofs into the $n$ Lab, we have both: no spoiler for those who want to figure it out themselves and at the same time eventually a nice colleciton of propositions and proofs.

For the present case, the solution is spelled out at $n$ Lab: kernel – Properties.

Posted by: Urs Schreiber on December 2, 2010 2:04 PM | Permalink | Reply to this

Re: Homological Algebra Puzzle

$MathML-enabled post (click for more details).$

Todd wrote:

The first “puzzle” is not hard; do I need to do the rot13 thing?

Nice solution! The only problem with making the solution easily visible is that almost nobody reading the puzzle can resist glancing down at the solution… and then they can’t really do the puzzle.

So if I write any more puzzles in the future, I think I’ll follow Urs’ advice and have people write answers in an nLab page that I link to. (This assumes that people are psychologically capable of ‘leaving the room’ — the blog, here — and putting the answer in another room. I’m not sure that’ll work, but we’ll see.)

The answer to the second question has to be “no”, because the solution to the puzzle has nothing to do with the fact that monomorphisms are kernels, etc…

Duh — I hadn’t even noticed that! Maybe I can try to salvage it: can we characterize abelian categories among those additive categories $A$ that have kernels and cokernels in terms of some appealing property of the adjoint functors

$coker : Arr(A) \rightarrow Arr(A)$ $Arr(A) \leftarrow Arr(A) : ker$

Posted by: John Baez on December 3, 2010 2:14 AM | Permalink | Reply to this

Re: Homological Algebra Puzzle

$MathML-enabled post (click for more details).$

Nothing terribly imaginative comes to mind. The $(ker, coker)$ adjunction restricts to an adjunction

$(Coker: Mono \to Epi) \dashv (Ker: Epi \to Mono)$

and if this adjunction is an equivalence (together with the additive category), we get an abelian category.

All this reminds me of a nice way of thinking about image factorizations (as in topos theory): if $X$ is an object of a category $E$ , the image (of a map targeted at $X$ ) provides a left adjoint to the full inclusion

$Sub(X) \to E/X$

Posted by: Todd Trimble on December 3, 2010 7:00 AM | Permalink | Reply to this

Re: Homological Algebra Puzzle

how about “the unit is epi and the counit is mono”?

the adjunction between cokernel and kernel in an abelian category is a sort of prelude to the inverse relationship between homotopy-cokernel and homotopy-kernel in the accompanying category of chain complexes. following todd’s observation, this inverse relationship can be seen as arising from the fact that all chain maps are “homotopy-monic” and “homotopy-epic”.

Posted by: james dolan on December 3, 2010 11:58 PM | Permalink | Reply to this

Re: Homological Algebra Puzzle

$MathML-enabled post (click for more details).$

Jim, I don’t believe “unit is epi, counit is mono” is enough to get abelianness if the additive category isn’t already balanced.

For example, in the additive category of locally compact Hausdorff abelian groups, I think the cokernel of $f: A \to B$ is the projection $q: B \to B/\widebar{f(A)}$ (mod out by the closure of the image of $f$ in $B$ ), and the kernel of this is the inclusion $\widebar{f(A)} \hookrightarrow B$ . The evident map $A \to \widebar{f(A)}$ is epic because it is dense, so I believe the unit is epic. Also, since this category is self-dual, the counit should be monic. But the category is not balanced (again consider the example where $f$ is the irrational line on a torus), and so is not abelian.

Posted by: Todd Trimble on December 4, 2010 1:37 AM | Permalink | Reply to this

The n-Category Café

Skip to the Main Content

December 2, 2010