Definition 4.3.4.1.1 (00BF): Pushouts of Pointed Sets

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The pushout of $(X, x_{0})$ and $(Y, y_{0})$ over $(Z, z_{0})$ along $(f, g)$ is the pair consisting of:

The Colimit. The pointed set $(X \underset{f, Z, g}{∐} Y, p_{0})$ , where:
- The set $X \underset{f, Z, g}{∐} Y$ is the pushout (of unpointed sets) of $X$ and $Y$ over $Z$ with respect to $f$ and $g$ ;
- We have $p_{0} = [x_{0}] = [y_{0}]$ .
The Cocone. The morphisms of pointed sets

$\begin{aligned} {inj}_{1} & : (X, x_{0}) \to (X \underset{Z}{∐} Y, p_{0}), \\ {inj}_{2} & : (Y, y_{0}) \to (X \underset{Z}{∐} Y, p_{0}) \end{aligned}$

given by

$\begin{aligned} {inj}_{1} (x) & \overset{def}{=} [(0, x)] \\ {inj}_{2} (y) & \overset{def}{=} [(1, y)] \end{aligned}$

for each $x \in X$ and each $y \in Y$ .

Firstly, we note that indeed $[x_{0}] = [y_{0}]$ , as we have

\begin{aligned} x_{0} & = f (z_{0}), \\ y_{0} & = g (z_{0}) \end{aligned}

since $f$ and $g$ are morphisms of pointed sets, with the relation $\sim$ on $X ∐_{Z} Y$ then identifying $x_{0} = f (z_{0}) \sim g (z_{0}) = y_{0}$ .

We now claim that $(X \underset{Z}{∐} Y, p_{0})$ is the categorical pushout of $(X, x_{0})$ and $(Y, y_{0})$ over $(Z, z_{0})$ with respect to $(f, g)$ in ${Sets}_{*}$ . First we need to check that the relevant pushout diagram commutes, i.e. that we have

Indeed, given

z \in Z

, we have

\begin{aligned} [{inj}_{1} \circ f] (z) & = {inj}_{1} (f (z)) \\ = [(0, f (z))] \\ = [(1, g (z))] \\ = {inj}_{2} (g (z)) \\ = [{inj}_{2} \circ g] (z), \end{aligned}

where $[(0, f (z))] = [(1, g (z))]$ by the definition of the relation $\sim$ on $X ∐ Y$ (the coproduct of unpointed sets of $X$ and $Y$ ). Next, we prove that $X ∐_{Z} Y$ satisfies the universal property of the pushout. Suppose we have a diagram of the form

in ${Sets}_{*}$ . Then there exists a unique morphism of pointed sets

ϕ : (X \underset{Z}{∐} Y, p_{0}) \to (P, *)

making the diagram

commute, being uniquely determined by the conditions

\begin{aligned} ϕ \circ {inj}_{1} & = ι_{1}, \\ ϕ \circ {inj}_{2} & = ι_{2} \end{aligned}

via

ϕ (p) = {\begin{cases} ι_{1} (x) & if x = [(0, x)], \\ ι_{2} (y) & if x = [(1, y)] \end{cases}

for each $p \in X \underset{Z}{∐} Y$ , where the well-definedness of $ϕ$ is proven in the same way as in the proof of Chapter 2: Constructions With Sets, Definition 2.2.4.1.1. Finally, we show that $ϕ$ is indeed a morphism of pointed sets, as we have

\begin{aligned} ϕ (p_{0}) & = ϕ ([(0, x_{0})]) \\ = ι_{1} (x_{0}) \\ = *, \end{aligned}

or alternatively

\begin{aligned} ϕ (p_{0}) & = ϕ ([(1, y_{0})]) \\ = ι_{2} (y_{0}) \\ = *, \end{aligned}

where we use that $ι_{1}$ (resp. $ι_{2}$ ) is a morphism of pointed sets.