---- > [!theorem] Theorem. ([[universal property of product sets]]) > Products $A \times B$ (or, more precisely, $A \times B$ together with information of $\pi_{A}$ and $\pi_{B}$) are [[terminal object|final]] in the [[two-object slice category|two-object slice]] [[category]] $\mathsf{Set}_{A,B}$. > > > Unpacking this: let $A,B$ be sets, and consider the product $A \times B$, with the two natural [[projection function|projections]] $\pi_{A}$ and $\pi_{B}$ thus: > > ```tikz > \usepackage{tikz-cd} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZARgBoAGAXVJADcBDAGwFcYkQBBEAX1PU1z5CKMgCZqdJq3YAhHnxAZseAkXKliEhizaJOAAgA6hvAFt4+udwkwoAc3hFQAMwBOEU0lE0cEJOskddmM0LAB9Ll4Xd09EbxBfJDIQRnoAIxhGAAUBFWEQVyw7AAscEBptaT0Q8KtKbiA > \begin{tikzcd} > & A \\ > A \times B \arrow[ru, "\pi_A"] \arrow[rd, "\pi_B"'] & \\ > & B > \end{tikzcd} > \end{document} > ``` > Then for any set $Z$ along with $f_{A}:Z \to A$ and $f_{B}:Z \to B$, there is a *unique* $\sigma:Z \to A \times B$ such that the following diagram commutes: > > ```tikz > \usepackage{tikz-cd} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBoBGAXVJADcBDAGwFcYkQAtEAX1PU1z5CKchWp0mrdgEEABAB15eALbxZAIR58QGbHgJEATKWLiGLNohDSt-PUKOlDZyZZCbu4mFADm8IqAAZgBOEMpIoiA4EEjGEhbsimhYAPo2vEGh4YiR0UgAzDTmUlZJqZo0jPQARjCMAAoC+sIgwVg+ABY4tiAhYUhkUTE5Ra6J8tg+yvQ9fdmDeYhxtWBQBYPFboFps1kDNIuFIFW1DU0OVm2d3aMJVtsVICtriAC0+cSe3EA > \begin{tikzcd} > & & A \\ > Z \arrow[r, "\sigma"] \arrow[rru, "f_A", bend left] \arrow[rrd, "f_B"', bend right] & A \times B \arrow[ru, "\pi_A"] \arrow[rd, "\pi_B"'] & \\ > & & B > \end{tikzcd} > \end{document} > ``` > > Specifically, $\sigma$ is the 'product function' $f_{A} \times f_{B}$. ^theorem > [!proof]- Proof. ([[universal property of product sets]]) > **Existence.** Define $\sigma:Z \to A \times B$ as $\sigma(z):=\big( f_{A}(z), f_{B}(z) \big)$. This manifestly makes the diagram commute. > **Uniqueness.** This is enforced by commutativity of the diagram. To convince myself: if we did not define $\sigma$ in this way, then there would be some $z$ with $\sigma(z) \neq (f_{A}(z), f_{B}(z))$, say WLOG that the first slot is not $f_{A}(z)$. Then $\pi_{A} \sigma(z) \neq f_{A}(z)$. ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` #reformatrevisebatch01