I am currently trying to make sense of products and coproducts in different categories, and even in starting with the basic examples (cartesian product and disjoint union in the category of sets) I've found that I don't understand how they meet the definition of a product and coproduct for one of the conditions.

The definition I was given is as follows. Given a functor $F:I\to C$ where $I$ is a discrete category and $C$ is some arbitrary category, a coproduct is an inductive limit of $F$ (a family of maps indexed by $I$ which is an initial object in the category of arrows ($F \to C$), similarly a product is the projective limit in the category ($C \to F$).

For a family of sets $(X_{i})_{i \in I}$, if we define a set $C$ that is their disjoint union, and equip it with a family of embedddings $\iota_{j} :X_{j} \hookrightarrow C$ by $x\mapsto (x,j)$ for each $j\in I$, I can see how it meets the other properties, but I still don't see why this object has to be initial for $(X_{i})_{i \in I}$.

What is it exactly that makes this family of mappings initial?

Thanks in advance.

  • $\begingroup$ The proof is one line long and does not need any ideas. Just write it down. $\endgroup$ – Martin Brandenburg Apr 14 '15 at 23:19
  • $\begingroup$ If I understood the concept I would have already done so. I'm asking for clarification. $\endgroup$ – Brandon Thomas Van Over Apr 14 '15 at 23:36

Hint 1: You are working with the universal property of limits in general. For most people, this certainly isn't the best starting point. Try writing out what the universal property means in this particular case (discrete domain) explicitly, it simplifies considerably. You can check what you got or (if you can't do it) what you're supposed to get by searching for the definition of (co)product, and not limit in general.

Hint 2: The universal property (the initiality condition you mention) consists of two parts: existence of a certain morphism, and it's uniqueness. Once you write everything out correctly, both steps are very short here, but it's worth keeping the distinction in mind in more complicated cases.

If this doesn't help, I'd recommend saying where exactly you are getting stuck, instead of asking about the entire process. It's easy enough for someone here to write it down for you, but you really better off doing it yourself.

  • $\begingroup$ That's what I was looking for. Thank you. $\endgroup$ – Brandon Thomas Van Over Apr 15 '15 at 0:49

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