# creation of products and coproducts

I'm wondering about the deduction of products and coproducts in $\bf Set$. They are both fairly simple objects, yet I cant find any constructive way to deduce them.

Tobias

Edit 1: Sorry for the vague question. I have a rather specific notion in mind. We can formulate the product of $J$ sets $A_j$ as the set of all maps $f:J\rightarrow \bigcup_{j\in J}A_j$ such that $f(j)\in A_j$ for all $j\in J$ with projections $p_i(f) = f(i)$

Dually, the coproduct can be formulated the union $\bigcup_{j\in J}(g:A_j\rightarrow J)$ such that $g(a)=j, \forall a\in A_j$ with injections $i_j(a) = (a,j)\in g$ for some $g$ in the coproduct

I like these formulations as they display the duality, are based on maps and seem constructive.

What I'd like to know is how to motivate their existence in the comma categories $(\bf{C}\overset{\triangle}{\rightarrow}\bf {C^J}\leftarrow \bf1)$ and its dual.

It feels like it should be simple, yet i've been stuck here for quite some time.

Edit 2: Motivation by adjunction would also be acceptable

Edit 3: So to clarify, I would like to motivate formulating the products and coproducts in $\bf Set$ as above by arguments based on objects in the comma category.

Again, apologies for not being clearer

Edit 4: Fixed definitions so others wondering the same might read it

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Do you mean proving they exist in general or computing them in specific cases? – Matthew Pressland Feb 22 '12 at 11:22
@user25470 I don't get it: do you want motivations for existance of products and coproducts in $\mathbf{Set}$ or in a comma category? If the latter could you write explicitly the functors defining said comma category? – Giorgio Mossa Feb 22 '12 at 13:16
Both your definitions of product and of coproduct are incorrect. – Arturo Magidin Feb 22 '12 at 17:26
TeX tip: when writing things like $\bigcup_{j\in J}A_j$ you want to use \bigcup as opposed to \cup, \bigcap and other such variants, known in the jargon as large operators or variable sized operators. – Mariano Suárez-Alvarez Feb 22 '12 at 21:51
@ Mariano Suárez-Alvarez;i dont really see the point, but ill keep it inmind – user25470 Feb 24 '12 at 10:54

I think I get what are you looking for: a presentation of the intuition that shows that those sets above are the product and coproduct of sets. Here's a way to do so. Consider a functor $\mathcal F \colon J \to \mathbf{Set}$ where $J$ is a discrete category. Let's consider a cone from the singleton set $*$ to $\mathcal F$, that is a natural transformation $\tau \colon \Delta^J(*) \to \mathcal F$, that is just a family $\langle \tau_i \colon * \to \mathcal F(i)\rangle_{i \in J}$. Let $\mathbf{Cone}(*,\mathcal F)$ be the set of cones from $*$ to $\mathcal F$, e consider the family of functions $$p_i \colon \mathbf{cone}(*,\mathcal F) \to \mathcal F(i)$$ defined as $p_i(\tau)=\tau_i(*)$. This is the limit of $\mathcal F$ i.e. the product of the $\{\mathcal F(i)\}_{i \in J}$, I omit the details here and invite you to read Mac Lane's Categories for working mathematicians. The point is that $\mathbf{Cone}(*,\mathcal F)$ is isomorphic, i.e. in bijection, with the usual cartesian product: infact we have that $\mathbf{Cone}(*,\mathcal F) = \prod_{i \in J}\mathbf{Set}(*,\mathcal F(i))$ and for every $i \in J$ we have that $\mathbf{Set}(*,\mathcal F(i)) \cong \mathcal F(i)$ in $\mathbf{Set}$, so we have an isomophism $i \colon \mathbf{Cone}(*,\mathcal F) \cong \prod_{i \in J} \mathcal F(i)$ and its easy to show that for every $i \in J$ the equality $\pi_i \circ i = p_i$ holds, where $\pi_i \colon \prod_{i \in J} \mathcal F(i) \to \mathcal F(i)$ is the ordinary projection from the cartesian product.

Your definition of the product is incorrect. It is not the union of all such maps, it is the set of all such maps. And the clause "$f(j)=a$ iff $a\in A_j$" is incorrect, since it would require that $f(j)=a$ for all $a\in A_j$ (which would mean that $f$ is not a map if $A_j$ has more than one element). Rather, the correct clause is "$f(j)\in A_j$ for all $j\in J$."
Likewise, your definition of the coproduct is incorrect; morally, it is right (you are trying to make the coproduct the set of all pairs $(a,j)$ with $a\in A_j$), but as given it is not well-defined if the sets $A_j$ are not pairwise disjoint. Instead, you want to define it as the union of all functions $\iota_j\colon A_j\to J$ defined by $\iota_j(a)=j$ for all $a\in A_j$ (that is, the union of the canonical immersions themselves).