The coproduct of $(X,x_{0})$ and $(Y,y_{0})$1 is the coproduct of $(X,x_{0})$ and $(Y,y_{0})$ in $\mathsf{Sets}_{*}$ as in 
, .
- 1Further Terminology: Also called the wedge sum of $(X,x_{0})$ and $(Y,y_{0})$.
6.3.3 Coproducts
Let $(X,x_{0})$ and $(Y,y_{0})$ be pointed sets.
Concretely, the coproduct of $(X,x_{0})$ and $(Y,y_{0})$, also called their wedge sum, is the pair consisting of:
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The Colimit. The pointed set $(X\vee Y,p_{0})$ consisting of:
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The Underlying Set. The set $X\vee Y$ defined by
where $\mathord {\sim }$ is the equivalence relation on $X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y$ obtained by declaring $(0,x_{0})\sim (1,y_{0})$.
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The Basepoint. The element $p_{0}$ of $X\vee Y$ defined by
\begin{align*} p_{0} & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[(0,x_{0})]\\ & = [(1,y_{0})]. \end{align*}
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The Cocone. The morphisms of pointed sets
\begin{align*} \mathrm{inj}_{1} & \colon (X,x_{0}) \to (X\vee Y,p_{0}),\\ \mathrm{inj}_{2} & \colon (Y,y_{0}) \to (X\vee Y,p_{0}), \end{align*}given by
\begin{align*} \mathrm{inj}_{1}(x) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[(0,x)],\\ \mathrm{inj}_{2}(y) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[(1,y)], \end{align*}for each $x\in X$ and each $y\in Y$.
Proof of Construction 6.3.3.1.2.
We claim that $(X\vee Y,p_{0})$ is the categorical coproduct of $(X,x_{0})$ and $(Y,y_{0})$ in $\mathsf{Sets}_{*}$. Indeed, suppose we have a diagram of the form
\[ \phi \colon (X\vee Y,p_{0})\to (C,*) \]
making the diagram
\begin{align*} \phi \circ \mathrm{inj}_{X} & = \iota _{X},\\ \phi \circ \mathrm{inj}_{Y} & = \iota _{Y} \end{align*}
via
\[ \phi (z)=\begin{cases} \iota _{X}(x) & \text{if $z=[(0,x)]$ with $x\in X$,}\\ \iota _{Y}(y) & \text{if $z=[(1,y)]$ with $y\in Y$} \end{cases} \]
for each $z\in X\vee Y$, where we note that $\phi $ is indeed a morphism of pointed sets, as we have
\begin{align*} \phi (p_{0}) & = \iota _{X}([(0,x_{0})])\\ & = \iota _{Y}([(1,y_{0})])\\ & = *, \end{align*}
as $\iota _{X}$ and $\iota _{Y}$ are morphisms of pointed sets.
Let $(X,x_{0})$ and $(Y,y_{0})$ be pointed sets.
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1.
Functoriality. The assignments
\[ (X,x_{0}),(Y,y_{0}),((X,x_{0}),(Y,y_{0}))\mapsto (X\vee Y,p_{0}) \]define functors
\begin{align*} X\vee - & \colon \mathsf{Sets}_{*} \to \mathsf{Sets}_{*},\\ -\vee Y & \colon \mathsf{Sets}_{*} \to \mathsf{Sets}_{*},\\ -_{1}\vee -_{2} & \colon \mathsf{Sets}_{*}\times \mathsf{Sets}_{*} \to \mathsf{Sets}_{*}. \end{align*} -
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Associativity. We have an isomorphism of pointed sets
\[ (X\vee Y)\vee Z\cong X\vee (Y\vee Z), \]natural in $(X,x_{0}),(Y,y_{0}),(Z,z_{0})\in \mathsf{Sets}_{*}$.
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Unitality. We have isomorphisms of pointed sets
\begin{align*} (\mathrm{pt},*)\vee (X,x_{0}) & \cong (X,x_{0}),\\ (X,x_{0})\vee (\mathrm{pt},*) & \cong (X,x_{0}),\end{align*}natural in $(X,x_{0})\in \mathsf{Sets}_{*}$.
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4.
Commutativity. We have an isomorphism of pointed sets
\[ X\vee Y \cong Y\vee X, \]natural in $(X,x_{0}),(Y,y_{0})\in \mathsf{Sets}_{*}$.
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Symmetric Monoidality. The triple $(\mathsf{Sets}_{*},\vee ,\mathrm{pt})$ is a symmetric monoidal category.
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The Fold Map. We have a natural transformation
called the fold map, whose component
\[ \nabla _{X} \colon X\vee X \to X \]at $X$ is given by
\[ \nabla _{X}(p)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} x & \text{if $p=[(0,x)]$,}\\ x & \text{if $p=[(1,x)]$} \end{cases} \]for each $p\in X\vee X$.
Proof of Proposition 6.3.3.1.3.
Item 1: Functoriality
This follows from , of .
Item 2: Associativity
Omitted.
Item 3: Unitality
Omitted.
Item 4: Commutativity
Omitted.
Item 5: Symmetric Monoidality
Omitted.
Item 6: The Fold Map
Naturality for the transformation $\nabla $ is the statement that, given a morphism of pointed sets $f\colon (X,x_{0})\to (Y,y_{0})$, we have 
\begin{align*} [\nabla _{Y}\circ (f\vee f)]([(i,x)]) & = \nabla _{Y}([(i,f(x))])\\ & = f(x)\\ & = f(\nabla _{X}([(i,x)]))\\ & = [f\circ \nabla _{X}]([(i,x)]) \end{align*}
for each $[(i,x)]\in X\vee X$, and thus $\nabla $ is indeed a natural transformation.
