Let $f\colon (X,x_{0})\to (Y,y_{0})$ be a morphism of pointed sets.
-
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.
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1.
Active-Inert Factorisation. Every morphism of pointed sets $f\colon (X,x_{0})\to (Y,y_{0})$ factors uniquely as
\[ f=a\circ i, \]where:
-
(a)
The map $i\colon (X,x_{0})\to (K,k_{0})$ is an inert morphism of pointed sets
-
(b)
The map $a\colon (K,k_{0})\to (Y,y_{0})$ is an active morphism of pointed sets.
Moreover, this determines an orthogonal factorisation system in $\mathsf{Sets}_{*}$.
-
(a)
-
•
$K$ is the pointed set given by
\begin{align*} K & = \left\{ x\in X\ \middle |\ f(x)\neq y_{0}\right\} \cup \left\{ x_{0}\right\} \\ & = (X\setminus f^{-1}(y_{0}))\cup \left\{ x_{0}\right\} ; \end{align*} -
•
$i\colon X\to K$ is the inert morphism of pointed sets given by
\[ i(x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} x & \text{if $x\in K$},\\ x_{0} & \text{otherwise} \end{cases} \]for each $x\in X$;
-
•
$a\colon K\to Y$ is the active morphism of pointed sets given by
\[ a(x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f(x) \]for each $x\in K$.
6.1.5 Active and Inert Morphisms of Pointed Sets
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.
Let $(X,x_{0})$ and $(Y,y_{0})$ be pointed sets.
Proof of Proposition 6.1.5.1.4.
Item 1: Active-Inert Factorisation
Let $f\colon X\to Y$ be a morphism of pointed sets. We can factor $f$ as 
Next, let
\[ \phi (y)=f(i^{-1}(y)) \]
for each $y\in Y$ (which is well-defined since, as $i$ is inert, $i^{-1}(y)$ is a singleton for all $y\in Y$). We claim that $\phi $ is the unique diagonal filler in the diagram
\begin{align*} [\phi \circ i](x) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\phi (i(x))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f(i^{-1}(i(x)))\\ & = f(x) \end{align*}
for each $x\in X$ and
\begin{align*} [a\circ \phi ](y) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}a(\phi (y))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}a(f(i^{-1}(y)))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[a\circ f](i^{-1}(y))\\ & = [g\circ i](i^{-1}(y))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g(i(i^{-1}(y)))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g(y) \end{align*}
for each $y\in Y$. Moreover, given another morphism $\psi $ such that the diagram
\begin{align*} \psi (y) & = \psi (i(x))\\ & = f(x)\\ & = f(i^{-1}(y))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\phi (y). \end{align*}
This finishes the proof.
