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Stream: theory: algebraic topology

Topic: distributing cokernels over pullbacks?

Benjamin Merlin Bumpus (he/him) (Feb 06 2024 at 16:25):

Hi all, this is perhaps a silly question, but I'm wondering: when can one say that a cokernels distribute over a pullback (in an Abelian category).

Specifically, suppose I have a map $f \colon Z \to L \times_M R$ into a pullback (see the diagram) when can we say that $\mathrm{coker}(p_1f) \times_{\mathrm{coker}(pf)} \mathrm{coker}(p_2f) = \mathrm{coker}f$ (where $p_1$, $p_2$ and $p$ are respectively the projections of the pullback into $L$, $R$ and $M$ respectively).

Morgan Rogers (he/him) (Feb 07 2024 at 13:08):

Using the universal properties of the cokernels, we get a universal map $\mathrm{coker}(f) \to \mathrm{coker}(p_1f) \times_{\mathrm{coker}(pf)} \mathrm{coker}(p_2f)$. I couldn't find a sensible way to reason about the reverse direction, though; given the context you would want to try to show that the kernel of that universal map is trivial, but I don't immediately see hypotheses guaranteeing that.

Morgan Rogers (he/him) (Feb 07 2024 at 13:09):

What kind of conditions do you have to work with?

Benjamin Merlin Bumpus (he/him) (Feb 07 2024 at 17:35):

@Zoltan A. Kocsis wrote me late at night a tentative proof that this is true if $L \times_M R$ is a pullback of monomorphisms of Abelian groups (he's an angel and wrote me before going to bed). I think the proof checks out, but neither he nor I have had the time to really make sure this works with a lucid mind.

Benjamin Merlin Bumpus (he/him) (Feb 07 2024 at 17:36):

I think the same argument can be extended to prove that the map
$\quad u \colon \mathrm{coker}f \to \mathrm{coker}(p_1f) \times_{\mathrm{coker}(pf)} \mathrm{coker}(p_2f)$
is an epimorphism.

However, I don't know under what conditions one can say that it is also a monomorphism...

Matteo Capucci (he/him) (Feb 08 2024 at 10:49):

One can reason about it with elements.
An element of ${\rm coker} f$ has form $(l,r) + {\rm im} f$ where $p_1l = p_2 r$. The comparison map $u$ sends such an element to the pair $(l + {\rm im} p_1f, r + {\rm im} p_2 f)$. For $u$ to be monic then $\ker u = 0$, which means $(l + {\rm im} p_1f, r + {\rm im} p_2 f) = 0$ iff $(l,r) \in {\rm im} f$ iff there is $z \in Z$ such that $f(z)=(l,r)$. This holds e.g. if $f$ is epi, but something weaker should be sufficient. In particular I think my argument shows $u$ is monic when $f$ is epi and $M=1$, which generalizes to a pullback for arbitrary $M$ but clearly there's more space for $u$ to be iso there.

Matteo Capucci (he/him) (Feb 08 2024 at 10:49):

I suspect the best way to approach this question is to arm yourself with patience and exact sequences and find the right one involving the kernel and cokernel of $u$

Matteo Capucci (he/him) (Feb 08 2024 at 10:50):

This might be helpful