Item 1 (01QE) — The Clowder Project

We write $\mathsf{Sets}^{\mathrm{actv}}_{*}$ for the wide subcategory of $\mathsf{Sets}_{*}$ spanned by pointed sets and the active maps between them.

Here are some examples of active and inert maps of pointed sets.

1.
The map $\mu \colon \left\langle 2\right\rangle \to \left\langle 1\right\rangle $ given by
is active but not inert.
2.
The map $f\colon \left\langle 2\right\rangle \to \left\langle 2\right\rangle $ given by
is inert but not active.
3.
The map $f\colon \left\langle 3\right\rangle \to \left\langle 1\right\rangle $ given by
is neither inert nor active. However, it factors as $f=a\circ i$, where

\begin{align*} i & \colon \left\langle 3\right\rangle \to \left\langle 2\right\rangle ,\\ a & \colon \left\langle 2\right\rangle \to \left\langle 1\right\rangle \end{align*}

are the morphisms of pointed sets given by
with $i$ being inert and $a$ being active.

Let $\webleft (X,x_{0}\webright )$ and $\webleft (Y,y_{0}\webright )$ be pointed sets.

1.
Active-Inert Factorisation. Every morphism of pointed sets $f\colon \webleft (X,x_{0}\webright )\to \webleft (Y,y_{0}\webright )$ factors uniquely as

\[ f=a\circ i, \]

where:
1. (a)
  The map $i\colon \webleft (X,x_{0}\webright )\to \webleft (K,k_{0}\webright )$ is an inert morphism of pointed sets
2. (b)
  The map $a\colon \webleft (K,k_{0}\webright )\to \webleft (Y,y_{0}\webright )$ is an active morphism of pointed sets.
Moreover, this determines an orthogonal factorisation system in $\mathsf{Sets}_{*}$.

Item 1: Active-Inert Factorisation

Let $f\colon X\to Y$ be a morphism of pointed sets. We can factor $f$ as

where:

•
$K$ is the pointed set given by

\begin{align*} K & = \left\{ x\in X\ \middle |\ f\webleft (x\webright )\neq y_{0}\right\} \cup \left\{ x_{0}\right\} \\ & = \webleft (X\setminus f^{-1}\webleft (y_{0}\webright )\webright )\cup \left\{ x_{0}\right\} ; \end{align*}
•
$i\colon X\to K$ is the inert morphism of pointed sets given by

\[ i\webleft (x\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} x & \text{if $x\in K$},\\ x_{0} & \text{otherwise} \end{cases} \]

for each $x\in X$;
•
$a\colon K\to Y$ is the active morphism of pointed sets given by

\[ a\webleft (x\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f\webleft (x\webright ) \]

for each $x\in K$.

Next, let

be a commutative diagram in $\mathsf{Sets}_{*}$. Consider the morphism $\phi \colon Y\to A$ given by

\[ \phi \webleft (y\webright )=f\webleft (i^{-1}\webleft (y\webright )\webright ) \]

for each $y\in Y$ (which is well-defined since, as $i$ is inert, $i^{-1}\webleft (y\webright )$ is a singleton for all $y\in Y$). We claim that $\phi $ is the unique diagonal filler in the diagram

Indeed, this diagram commutes, as we have

\begin{align*} \webleft [\phi \circ i\webright ]\webleft (x\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\phi \webleft (i\webleft (x\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f\webleft (i^{-1}\webleft (i\webleft (x\webright )\webright )\webright )\\ & = f\webleft (x\webright ) \end{align*}

for each $x\in X$ and

\begin{align*} \webleft [a\circ \phi \webright ]\webleft (y\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}a\webleft (\phi \webleft (y\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}a\webleft (f\webleft (i^{-1}\webleft (y\webright )\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [a\circ f\webright ]\webleft (i^{-1}\webleft (y\webright )\webright )\\ & = \webleft [g\circ i\webright ]\webleft (i^{-1}\webleft (y\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g\webleft (i\webleft (i^{-1}\webleft (y\webright )\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g\webleft (y\webright ) \end{align*}

for each $y\in Y$. Moreover, given another morphism $\psi $ such that the diagram

commutes, it follows that we must have $\psi =\phi $, since, given $y\in Y$, there exists a unique $x\in X$ such that $i\webleft (x\webright )=y$, so we have

\begin{align*} \psi \webleft (y\webright ) & = \psi \webleft (i\webleft (x\webright )\webright )\\ & = f\webleft (x\webright )\\ & = f\webleft (i^{-1}\webleft (y\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\phi \webleft (y\webright ). \end{align*}

This finishes the proof.

6.1.5 Active and Inert Morphisms of Pointed Sets