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Stream: learning: questions

Topic: Non-commutation of finite limits and non-filtered colimits

David Michael Roberts (Jun 18 2023 at 04:01):

I'm wondering what is an example of a diagram in $\mathbb{Ab}^{\to}$, of a diagram that is non-filtered so that the kernel of the colimit is not the colimit of the kernels. Presumably this is easiest if one of them is 0, and one is not.

David Michael Roberts (Jun 18 2023 at 04:08):

This is in the context of the a paper that in a previous version tried working with the colimit over the Čech cochain complexes directly, instead of taking cohomology first, and then taking the colimit. The authors have since stopped using this, but I was really curious to see what actually goes wrong. The cryptic hint was that the (opposite of the) diagram of open covers wasn't filtered, so it's clearly to to do with the interaction of limits and colimits, but the explanation of the resulting problem seemed to me to be a bit spurious. Something about a certain item not being well-defined, but I'm not sure the specific item in question is the problem.

David Michael Roberts (Jun 18 2023 at 06:11):

I'm aware that with care, in essence via a kind of rigidification of the cochain complexes, one can in fact take the colimit and get the right thing on taking cohomology. For instance by ordering the topological space, or by considering special kinds of open covers and restricting what the refinement maps could be.

But there were none of those techniques coming to play, so I'm really just trying to understand the general case, where it breaks.

I guess is comes down to the ordinary colimit not calculating the homotopy colimit of cochain complexes correctly. But taking the colimit after cohomology means that one has already passed through a homotopy category, and so the colimit is less wrong?

Reid Barton (Jun 18 2023 at 07:51):

How about the span $\mathbb{Z} \leftarrow 0 \rightarrow \mathbb{Z}$ mapping to the span $\mathbb{Z} \leftarrow \mathbb{Z} \rightarrow \mathbb{Z}$, where all the maps $\mathbb{Z} \rightarrow \mathbb{Z}$ are the identity, viewed as a span in the arrow category. All the arrows are injective so their kernels are zero, but the colimit arrow $\mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$ is not injective (it's the sum map).

David Michael Roberts (Jun 18 2023 at 10:55):

Yeah, I thought something like that might work. But for covers, you have more in the diagram, in the sense that every cospan in the opposite of the category of open covers, there's a cocone, which isn't the case in the example. It needs to use the lack of cocones for parallel pairs, somehow.

David Michael Roberts (Jun 19 2023 at 00:55):

Hmm, I had a think, and came up with another example, this time of a coequaliser $(0\to \mathbb{Z}) \rightrightarrows (\mathbb{Z} \stackrel{id}{\to} \mathbb{Z})$ where one arrow is just inclusion and the other is multiplication by 2. These even have the property that they are chain homotopic, if I haven't messed things up, which is what happens in the case of Čech cohomology. The coequaliser can be calculation as the cokernel of the difference, which is then $\mathbb{Z}\to 0$, but as in your example, the kernels are both trivial.