# Intuition on product definition in category theory

I just have a question on product of 2 objects in category theory.

Given a category $$\mathcal{C}$$ we would like to define a product of 2 objects $$A,B \in \mathcal{C}$$. The definition says that $$D \in \mathcal {C}$$ is a product of $$A$$ and $$B$$ if for all $$E\in \mathcal{C}$$ there exists a unique morphisme $$m:C\to D$$ such that $$p\circ m=f$$ and $$q\circ m=g$$ with $$p:D\to A$$, $$q:D\to B$$, $$f:E\to A$$ and $$g:E\to B$$.

I would like to understand why this definition allows us for real to say that $$D$$ is a product of $$A$$ and $$B$$. I understand that given the triplet $$(p,q,D)$$ is not enough to say that $$D$$ is a "good" product; so we need to add some conditions on projections and $$D$$. Those conditions are exactly the unicity of morphism $$m$$ and applications $$f,g$$. But, I don't really see why adding these conditions allow to define the product of 2 objects "correctly". If someone could clarify all of this and give his intuition on this, I would really appriciate it. Thank you in advance

• What properties does the cartesian product $A \times B$ of two sets have? How about the direct product of groups? What's common is your definition. Feb 4 at 13:39
• @Randall I think I'm starting to see what's going on. If I keep in mind the cartesian product in cat $\textbf{Sets}$ and use the same notations as in the post, I need like to "redefine" the product $D$ by arrows and so for this I need an auxiliary object $E$ which has as well applications to $A$ and $B$ because I know that $D$ is a "composition" of elements of $A$ and $B$. Is it correct? But now, why the morphism $E\to D$ must be unique? And so, if I reformulate, why the product might be necessarily a terminal object ? Feb 4 at 13:52
• By putting "with $p:D\to A, q:D\to B, f:E\to A$ and $g:E\to B$" all at the end of your "definition" of product, you've lumped $p,q,f,g$ together in a way that confuses the logical structure of the definition. You should go back to the actual definition and understand that $p$ and $q$ are, along with $D$, part of the structure of the product, while $f$ and $g$ are, along with $E$, universally quantified in the definition. Feb 4 at 16:51

Here is the short answer: The conditions which you describe make sure that $$\mathcal C(z,a\times b) \to \mathcal C(z,a) \times \mathcal C(z,b)$$, $$f \mapsto (\pi_af,\pi_bf)$$ is an isomorphism for all $$z$$. This is why you can say sentences like: A map into $$a\times b$$ is the same as a pair of maps (one into $$a$$ and one into $$b$$).
Here is the long version. I assume that you know about the Yoneda lemma. A lot of interesting definitions in category theory are actually representations of $$\mathtt{Set}$$-valued functors. A representation of a contravariant functor $$F: \mathcal C \to \mathtt{Set}$$ is an object $$c$$ in $$\mathcal C$$ together with a natural isomorphism $$\alpha: \mathcal C(-,c) \to F$$ of functors. A representation usually tells us that the maps into $$c$$ have a nice description, namely $$F$$.
For example consider $$\mathcal C = \mathtt{Set}$$ and $$c = \{0,1\} = \Omega$$. Then the set maps from $$A$$ into $$\Omega$$ correspond naturally to the subsets of $$A$$, and $$\mathtt{Set}(-,\Omega)$$ is naturally isomorphic to the power set functor.
There are many (!) examples of representable covariant and contravariant functors in category theory. You could define the product of two objects $$a$$ and $$b$$ as follows. It should be an object $$a\times b$$ such that maps into $$a\times b$$ correspond naturally to pairs of maps into $$a$$ and $$b$$. In other words you like to have a representation $$\alpha: \mathcal C(-,a\times b) \to \mathcal C(-,a) \times \mathcal C(-,b)$$. This determines $$a\times b$$ up to unique compatible isomorphisms.
Now here is the catch. Because of Yoneda's lemma giving a natural transformation $$\alpha: \mathcal C(-,c) \to F$$ is the same as specifying an element $$\alpha(1_c)\in Fc$$, and $$\alpha$$ is an isomorphism if and only if $$\alpha(1_c) \in Fc$$ is terminal in the category of elements of $$F$$. Now in the case of a representation $$\mathcal C(-,a\times b) = \mathcal C(-,a) \times \mathcal C(-,b)$$ this means that specifying the natural isomorphism is the same as giving a pair of morphisms $$(\pi_a,\pi_b) \in \mathcal C(a\times b,a)\times \mathcal C(a\times b,b)$$. The condition that the transformation is an isomorphism becomes the universal property of the product, which you describe in your question.