The Factorization of Functors + Update

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

Let \text{Seq} be a posetal category, $[n]$ be a finite chain category, and \text{Ab} be an abelian category. Then let there be a functor F:\text{Seq}\to\text{Ab} such that

F(n)=\{\text{free abelian group on} \ n \ \text{generators}\}

if n\leq10 and

F(n)=\{\text{free abelian group on} \ 10 \ \text{generators}\}

if n>10.

We take the morphism [n]\to[n+1] to either the subgroup inclusion of the free group on n-generators into the free group on n+1-generators (in the case that n+1 is less than or equal to 10), and the identity otherwise. So so far, we have defined a functor F:\text{Seq}\to\text{Ab} by telling what each object and achieve morphism go to (and you can check that F preserves compositions of morphisms).

By factoring [n] into F, we are given a functor G:\text{Seq}\to[10], that takes \{1\mapsto1, 2\mapsto2, \cdots, 9\mapsto9, 10\mapsto10, 11\mapsto10, 12\mapsto10, \cdots\} is formed, and another functor H:[10]\to\text{Ab} the takes k\in\{1,\cdots,10\} to the free abelian group on k-generators.

From this, we are given

\text{Seq}\rightarrow^G[10]\rightarrow^H\text{Ab}.

And therefore, when taking a look at what F looks like in comparison to the factorization, you will see that H\circ G=F.

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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