Products of Objects

Products

Product of Objects

In a category $\mathbf{C}$, a product for the objects $A$ and $B$ consists of an object $P$ and arrows $p_1:P\to A$ and $p_2:P\to B$: satisfying the following universal mapping property: For all $X$ with $x_1$ and $x_2$, there exists unique $u:X\to P$ such that the following diagram commutes: We write such product $P$ as $A\times B$, and $⟨x_1,x_2⟩$ for $u:X\to A\times B$.

e.g.

• Products in $\mathbf{Set}$ is Cartesian products.
• Products in $\mathbf{Pos}$ is the greatest lower bound (infimum) as we require that $p\times p\le p$, $p\times q\le q$ and $x\le p,x\le q⟹x\le p\times q$.
• Products in $\mathbf{Top}$ is the product topology.

Proposition

Products are unique up to isomorphism.

Proof Suppose we have $P$ and $Q$ are products of $A,B\in \text{obj}\mathbf{C}$: Then there is unique $u:P\to Q$ such that $q_2\circ u=p_2$ and $q_1\circ u=p_1$. Similarly, exists unique $w:Q\to P$ such that $p_1\circ w=q_1$ and $p_2\circ w=q_2$. Therefore,$q_1\circ u\circ w=q_1,p_1\circ w\circ u=p_1$It follows that $u\circ w=1_Q$ and $w\circ u=1_P$, thus $P\cong Q$.

Proposition

Prop The binary product of objects is associative up to isomorphism:$(A\times B)\times C\cong A\times (B\times C)$

Def All Finite Products A category $\mathbf{C}$ is said to have all finite products if it has a terminal object and all binary products. The category $\mathbf{C}$ has all (small) products if every set of objects in $\mathbf{C}$ has a product.

Duality Principle

Formal Duality

For any statement $\Sigma$ in the language of category theory, if $\Sigma$ follows from the axioms for categories, then so does ${\Sigma }^{\ast }$:$\mathbf{C}⊢\Sigma ⟹\mathbf{C}⊢{\Sigma }^{\ast }$

Conceptual Duality

For any statement $\Sigma$ about categories, if $\Sigma$ holds for all categories, then so does the dual statement $\Sigma$.

Proof If $\Sigma$ holds for all categories $\mathbf{C}$, then it also holds in all categories ${\mathbf{C}}^{\text{op}}$, but then ${\Sigma }^{\ast }$ holds in all categories $({\mathbf{C}}^{\text{op}}{)}^{\text{op}}$, thus in all categories $\mathbf{C}$. $\square$

Coproducts

Definition

Def Coproduct In any category $\mathbf{C}$, $Q$ is a coproduct (dual product) of $A$ and $B$ if for any $Z$ and $z_1:A\to Z$ and $z_2:B\to Z$, there is a unique $u:Q\to Z$ with $u\circ q_i=z_i$ for $i=1,2$, all as indicated in: We usually write $Q=A+B$ for the coproduct, and $[z_1,z_2]$ for the uniquely determined morphism $u:A+B\to Z$. And $q_1,q_2$ are usually called coproduct injections.

e.g.

• In $\mathbf{Sets}$, the coproduct $A+B$ of two sets is their disjoint union $A+B=\{(a,A)\mid a\in A\}\cup \{(b,B)\mid b\in B\}$ with evident coproduct injections $i_1(a)=(a,A)$ and $i_2(b)=(b,B)$. And we we have $[z_1,z_2](x,\delta )=\{\begin{array}{ll}{z}_1(x)& \delta =1\\ z_2(x)& \delta =2\end{array}$
• In $\mathbf{Sets}$, every finite set $A$ is a coproduct: $A\cong 1+1+\cdots +1(\text{car}A\text{ times})$This is because a function $f:A\to Z$ is uniquely determined by its values $f(a)$ for all $a\in A$.
• If $M(A)$ and $M(B)$ are free monoids on sets $A$ and $B$, then in $\mathbf{Mon}$ we can construct their coproduct as $M(A)+M(B)\cong M(A+B)$.
• In a fixed poset $\mathbf{Pos}$, a coproduct of two elements $p,q$ is the supremum of $p$ and $q$.

Proposition

Prop A coproduct of two objects is exactly their product in the opposite category. Proof It is clear by definition.

Def Free Product In $\mathbf{Groups}$, given groups $A$ and $B$, the free group of words $G(|A|+|B|)/\sim$ is usually called the free product of $A$ and $B$, written $A\oplus B$. Prop The free product is actually a coproduct.

Prop In the category $\mathbf{Ab}$ of abelian groups, the coproduct of $A$ and $B$ is the cartesian product of the underlying sets $A$ and $B$.

Prop In the category $\mathbf{Ab}$ of abelian groups, there is a canonical isomorphism between the binary coproduct and product, $A+B\cong A\times B$Proof Let $\pi =[⟨1_A,0_B⟩,⟨0_A,1_B⟩]:A+B\to A\times B$. Then for all $(a,b)\in A+B$, $\begin{array}{rl}\pi (a,b)& =[⟨1_A,0_B⟩,⟨0_A,1_B⟩](a,b)\\ & =⟨1_A,0_B⟩(a)+⟨0_A,1_B⟩(b)\\ & =(1_A(a),0_B(a))+(0_A(b),1_B(b))\\ & =(a,0_B)+(0_A,b)\\ & =(a+0_A,0_B+b)\\ & =(a,b)\end{array}$

Definition

Def Coproduct Functor Since we define the coproduct of two morphisms as $f+f^{\prime }:A+A^{\prime }\to B+B^{\prime }$which leads to a coproduct functor $+:\mathbf{C}\times \mathbf{C}\to \mathbf{C}$ on categories $\mathbf{C}$ with binary coproducts.

Prop Coproducts are unique up to isomorphism. Proof By duality, and the fact that the dual of an isomorphism is an isomorphism.

Prop Coproducts are associative. That is $(A+B)+C\cong A+(B+C)$. Proof By duality, and the fact that the dual of an isomorphism is an isomorphism.