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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.

Please help

Thanks in advance,

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? –  Matt 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
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2 Answers 2

up vote 1 down vote accepted

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.

Hope this completely answers your question.

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Thx, this is exactly what i was looking for. I apologise for all sloopy formulations. –  user25470 Feb 24 '12 at 10:44
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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).

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Very true and thx for sorting that out. But how does one derive these object in a constructive way? –  user25470 Feb 22 '12 at 20:01
    
@user25470: Like others, I am puzzled by just what it is you are looking for; to me, the explicit definition is already a "deriv[ation] of these objects in a constructive way." We are giving, explicitly, a construction of these objects. So, whatever it is you are meaning to ask, it's not what you are actually asking –  Arturo Magidin Feb 22 '12 at 20:20
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