## 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.