Categorical products and coproducts · DL Ferrario's Test Web Page

# Categorical products and coproducts

## Universal property of coproducts

Let $X_1, X_2$ be objects of the category $\vC$. The coproduct of $X_1$ and $X_2$ (if it exists) is a space $X_1 \coprod X_2$ and two morphisms $j_1 \from X_1 \to X_1 \coprod X_2$, $j_2 \from X_2 \to X_1 \coprod X_2$ such that the following diagram commutes (the first is with MathJax AMScd, the second is an embedded SVG produced with xypic):

$\require{AMScd}$ \begin{CD} X_1 @>{j_1}>> X_1 \coprod X_2 @<{j_2}<< X_2 \\
@| @VVhV @| \\
X_1 @>{h_1}>> Z @<{h_2}<< X_2 \\
\end{CD}

Given a set $S$ of indices, the coproduct of a family $X _ {s}$, $s\in S$, of objects is (if it exists) the space $\coprod _ {s \in S} X _ s$, together with morphisms $j _ \alpha \from X _ \alpha \to \coprod _ {s \in S} X_ s$, such that for each $Z$ and each family of morphisms $h_ \alpha \from X_ \alpha \to Z$ there exists a unique $h\from \coprod_ {s \in S} X_ s \to Z$ such that for each $\alpha \in S$ on has $hj_ \alpha = h_ \alpha$, i.e. the following diagram commutes:

(1)
The coproduct in $\mathbf{Top}_*$ is the wedge of the pointed spaces.

Proof: For each $\alpha \in S$, let $x_ \alpha \in X_ \alpha$ denote the base-point of $X_ \alpha$. Let $\bigvee_ {s \in S} X_ s$ denote the disjoint union of all $X_ \alpha$, quotient after the equivalence relation with a single non-trivial equivalence class with all base-points $x_ s$, and with the quotient topology.
Given $h_ \alpha$, pointed maps, $h_ \alpha (x_ \alpha) = z_ 0$, where $z_ 0$ is the base-point of $Z$. Hence the map defined on the disjoint union (which is continuous) induces a (continuous) map on the quotient $h \from \bigvee_ {s \in S} X_ s \to Z$. (/Proof)

## Products

Reversing arrows, one define the product, denoted by $\prod_ {s \in S} X_ s$, together with projections $p_ \alpha \from \prod_ {s \in S} X_ s \to X_ \alpha$, in any category $\mathbf{C}$, with the same universal property.

By definition, the product topology in $\mathbf{Top}$ is the topology on the categorical product, i.e. the topology on $\prod_ {s \in S} X_ s$ such that a function $f\from Z \to \prod_ {s \in S} X_ s$ is continuous if and only if its components $f_ \alpha = p_ \alpha f$ are continuous.

One can show that the product topology is generated by the base of all subsets of type $\prod_ {s \in S} U_ s \subset \prod_ {s \in S} X_ s$, where for each $\alpha\in S$ the subspace $U_ \alpha \subset X_ \alpha$ is an open subset of $X_ \alpha$, and for all but a finite number of $\alpha$'s the equality $U_ \alpha = X_ \alpha$ holds. The topology generated by subsets $\prod_ {s \in S} U_ s$ without the finiteness condition is the box topology, which is not equivalent to the product topology.

(2)
The product in $\mathbf{Top}_ *$ is the cartesian product, with the product topology. The base point of $\prod_ {s \in S } (X_ s, \hat x_ s)$ is $(\hat x_ s)_ {s \in S}$.

Proof: Let $(x_ s)_ {s \in S}$ denote a generic element of the product $\prod_ {s \in S } (X_ s, \hat x_ s)$, with $x_ s \in X_ s$ for each $s \in S$. Given maps $h_ \alpha$, the map $h$ defined by $h(z) = (h_ \alpha(z))_ {\alpha\in S}$ is well-defined and continuous in $\mathbf{Top}$. Furthermore, if $\hat z\in Z$ is the base-point, then $h_ \alpha(\hat z) = \hat x_ \alpha$ for each $\alpha$, hence $h(\hat z)$ is the base-point in $\prod_ {\alpha \in S } (X_ \alpha, \hat x_ \alpha)$. (/Proof)

If $X_ \alpha$ does not depend upon $\alpha$, then the product $\prod_ {\alpha \in S} X_ \alpha$ is also denoted as $X^S$ and is the space of all functions $S \to X$. The projections $p_ \alpha \from X^S \to X$ are therefore the evaluations $x \in X^S \mapsto x(\alpha) \in X$. Hence a sequence $x_ n$ of functions $S \to X$ converges in the product topology if and only if for each projection $p_ \alpha$ the projection $p_ \alpha(x_ n)$ converges, which means that $x_ n(\alpha)$ converges for each $\alpha$, or that $x_n$ converges pointwise.

## Free abelian groups

The free abelian group (i.e., $\ZZ$-module) generated by the set $S$ is the coproduct in $\mathbf{Ab}$ $$\ZZ [ S ] = \coprod_ {\alpha \in S} \ZZ = \bigoplus_ { \alpha \in S} \ZZ ,$$ since the coproduct in $\mathbf{Ab}$ is the direct sum $\bigoplus$.

The free (non-abelian) group generated by the set $S$ is $$F(S) = \coprod_{\alpha \in S} \ZZ = \ZZ * \ZZ * \cdots$$ where the coproduct is taken in the category of groups $\mathbf{Grp}$, and hence it is the free product $*$.

## Posets

A poset (Partially Ordered Set) is a category $\mathbf{P}$ in which hom-sets have either 0 or 1 element, and such that if there exist morphisms $A\to B$ and $B\to A$, then $A=B$.

So, if $A$ and $B$ are objects one can define [ A \leq B \iff \mathbf{P}(A,B) = \hom_{\mathbf{P}}(A,B) \neq \emptyset ]

(3)
The relation $A \leq B$ in $\mathbf{P}$ is reflexive, antisymmetric and transitive.

Proof: Since $1_A \in \hom(A,A)$, the inequality $A\leq A$ holds. If $A\leq B$ and $B\leq A$, then there are morphisms $A\to B$ and $B \to A$, and by assumpion $A=B$. Transitivity follows from the existence of composition of morphisms in any category. (/Proof)

(4)
The coproduct in $\mathbf{P}$ of a family $X_ \alpha$ of objects is (if it exists) the supremum $\sup_ {s \in S} (X_ s)$ in $\mathbf{P}$; the product in $\mathbf{P}$ of a family $X_ \alpha$ of objects is (if it exists) the infimum $\inf_ {s \in S} (X_ s)$.

Proof: Because of the $j_ \alpha \from X_ \alpha \to \coprod_ {s \in S} X_ s$, the coproduct is an upper bound, i.e. $X_ \alpha \leq \coprod_ {s \in S} X_ s$ for each $\alpha \in S$. If $Z$ is any other upper bound, i.e. an object such that for each $\alpha \in S$ the inequality $X_ \alpha \leq Z$ holds, then there exists $h \from \coprod_ {s \in S} X_ s \to Z$, and hence $\coprod_ {s \in S} X_ s \leq Z$, i.e. the coproduct is the minimum of the upper bounds.

Reversing arrows in $\mathbf{P}$ one obtains the result for the product. (/Proof)


Example: Consider the poset $\bP$ of all subsets of a set $S$, with $A\leq B$ if and only if $A\subseteq B \subseteq S$. Then the coproduct and the product are the union and intersection of the sets, respectively: $$\coprod_ {\alpha} X_ \alpha = \bigcup_ \alpha X_ \alpha$$ $$\prod_ {\alpha} X_ \alpha = \bigcap_ \alpha X_ \alpha$$

## Coproducts for the Working Mathematician

Categories for the Working Mathematician is the go-to reference for anything categorical. Let $\bC, \bD$ be categories, and $F\from \bC \to \bD$ a functor. The comma category $X \downarrow F$ of objects $F$-under $X$, for $X$ object in $\bD$, has as objects the pairs $(f,Y)$ with $Y\in \bC$ and $f\from X \to F(Y)$, and as morphisms (arrows) $h \from (f,Y) \to (f',Y')$ all morhpisms $h\from Y \to Y'$ su h that $f'=F(h) f$ as in the following diagram.

Given $X\in \bD$ and $F\from \bC \to \bD$, the comma category $X \downarrow F$ is uniquely defined. If $X\downarrow F$ has an initial object, then this is termed the universal arrow from $X$ to $F$.

(5)
The coproduct of two objects $X_ 1,X_ 2$ in $\bC$ is the initial object of the comma category $(X_ 1,X_ 2) \downarrow \Delta$, where $\Delta\from \bC \to \bC\times \bC$ is the diagonal functor. Hence it is the universal arrow from $(X_1,X_2)$ to $\Delta$.

Proof: Consider the diagonal functor $F=\Delta \from \bC \to \bD = \bC \times \bC$ (which sends $Y$ to $(Y,Y)$ for each object $Y$ of the category $\bC$, and morphisms accordingly). Given an element $X=(X_ 1,X_ 2) \in \bD = \bC\times \bC$, the initial object (if it exists) of the comma category $(X_ 1,X_ 2) \downarrow \Delta$ is exactly the coproduct $X_ 1 \coprod X_ 2$ in $\bC$. In fact, an initial object is an object $(j,A) \in (X_ 1,X_ 2) \downarrow \Delta$ such that for each object $(h,Z) \in (X_ 1,X_ 2) \downarrow \Delta$ there exists a unique morphism $(j,A) \to (h,Z)$. In other words, an initial object is a pair $(j,A)$ with $A\in \bC$ and $j \from (X_ 1, X_ 2) \to (A,A)$ such that for each $(h,Z)$, with $h\from (X_ 1, X_ 2) \to (Z,Z)$ there exists a unique morphism $f \from (j,A) \to (h,Z)$, i.e. a unique morphism $f \from A \to Z$ such that the composition $\Delta(f) j\from (X_ 1,X_ 2) \to (A,A) \to (Z,Z)$ is equal to $h$. (/Proof)