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 & = \webleft\{ x\in X\ \middle |\ f\webleft (x\webright )\neq y_{0}\webright\} \cup \webleft\{ x_{0}\webright\} \\ & = \webleft (X\setminus f^{-1}\webleft (y_{0}\webright )\webright )\cup \webleft\{ x_{0}\webright\} ; \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.
