What is dual to Cartesian Product

Question

I know that "Cartesian Product" of two sets $A$ and $B$ is the set of all possible tuples of $A$ and $B$. What is the dual of Cartesian Product? $\left(\frac AB\right)$? And how does it work?

Answer 1

In the sense of categorical duality, the dual of the cartesian product is the disjoint union. This is because:

• The product $A \times B$ is characterised by the fact that, for all sets $X$, functions $X \to A \times B$ correspond with pairs of functions $X \to A$ and $X \to B$.
• The disjoint union $A \sqcup B$ is characterised by the fact that, for all sets $X$, functions $A \sqcup B \to X$ correspond with pairs of functions $A \to X$ and $B \to X$.

To see this, we have to characterise the cartesian product in terms of its universal property—specifically, given two sets $A$ and $B$, their cartesian product is (up to bijection) the set $P$ equipped with projection functions $A \xleftarrow{\pi_A} P \xrightarrow{\pi_B} B$ such that, for all functions $A \xleftarrow{f} X \xrightarrow{g} B$, there is a unique function $u : X \to P$ such that $\pi_1 \circ h = f$ and $\pi_2 \circ h = g$. $$\begin{matrix} && X && \\ & {}^f\swarrow\phantom{{}^{f}} & {\tiny \phantom{h}}\downarrow{\tiny h} & \phantom{{}^g}\searrow^g& \\ A & \underset{\pi_A}{\leftarrow} & P & \underset{\pi_B}{\rightarrow} & B \end{matrix}$$ We do this by taking $P = A \times B = \{ (a,b) \mid a \in A, b \in B \}$ and $\pi_1(a,b)=a$ and $\pi_2(a,b)=b$. The unique morphism $h$ arising from $f,g$ is defined by $h(x)=(f(x),g(x))$ for all $x \in X$.

The dual is obtained by reversing all the arrows. That is, it must be a set $C$ equipped with functions $A \xrightarrow{\iota_A} C \xleftarrow{\iota_B} B$ such that, for all functions $A \xrightarrow{f} X \xleftarrow{g} B$, there is a unique function $h : C \to X$ such that $h \circ \iota_A = f$ and $h \circ \iota_B = g$. $$\begin{matrix} && X && \\ & {}^f\nearrow\phantom{{}^{f}} & {\tiny \phantom{h}}\uparrow{\tiny h} & \phantom{{}^g}\nwarrow^g& \\ A & \underset{\iota_A}{\rightarrow} & C & \underset{\iota_B}{\leftarrow} & B \end{matrix}$$ It turns out that taking $C$ to be the disjoint union $A \sqcup B$ of $A$ and $B$, and letting $\iota_A$ and $\iota_B$ be the respective inclusions of $A$ and $B$ into the disjoint union, satisfies the desired universal property. The function $h$ is defined by letting $$h(x) = \begin{cases} f(x) & \text{if } x \in A \subseteq A \sqcup B \\ g(x) & \text{if } x \in B \subseteq A \sqcup B \end{cases}$$