(Huge thanks to Jon Beardsley for this wonderful discussion.)

Let be a posetal category, $[n]$ be a finite chain category, and be an abelian category. Then let there be a functor such that

if and

if .

We take the morphism to either the subgroup inclusion of the free group on -generators into the free group on -generators (in the case that is less than or equal to ), and the identity otherwise. So so far, we have defined a functor by telling what each object and achieve morphism go to (and you can check that preserves compositions of morphisms).

By factoring into , we are given a functor , that takes is formed, and another functor the takes to the free abelian group on -generators.

From this, we are given

.

And therefore, when taking a look at what looks like in comparison to the factorization, you will see that .

—

My apologies for no post yesterday as planned. I am currently on the road going to *some place. *Which* *puts me in a position to be typing this post on my phone in a very cramped vehicle. However that is not stopping me from fulfilling my two-post-every-other-week plan.

A major update that I would like to make is that Mondays will now be a discussion on what I learned and what I thought about in reading Munkres’s *Topology.*

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