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Awodey states, at the end of page 47 of his 2010 book (where he discusses products), the following:

[...] one can define the product of a family of objects $(C_i)_{i \in I}$ indexed by any set $I$, by giving a UMP for '$I$-ary products' [...]. We leave the precise formulation [...] as an exercise.

Now this would be my approach (see below), and I would appreciate if anyone has any comment. I am struggling a bit with the UMP, although in this case it seems that just re-writing the UMP for binary products, should be the right way to define the '$I$-ary prododuct', as Awodey calls it.

Consider a category $\mathbf{C}$ and an indexed family of objects $(C_i)_{i \in I}$ in said category. The $I$-ary product for this category consists of an object $P=\prod_{i\in I} C_i$ and arrows $p_i: P \to C_i$ for all $i\in I$, which satisfy the following UMP: Given $x_i:X_i\to C_i$ were $X_i$ is a member of an index family of objects $(X_i)_{i\in I}$ in $\mathbf{C}$, there exists a unique $u = \langle u_{i}\rangle_{i\in I} :X \to P$, with $X=\prod_{i\in I} X_i$ such that $p_i\circ u_i=x_i$, for all $i\in I$.

Would this be correct/sufficient to define the UMP for $I$-ary products?

(Drawing commutative diagrams or including a picture of them would have been easier, but AMScd does not support diagonal arrows.)

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  • $\begingroup$ $I$ may not contain elements $1$, $2$, $3,\ldots$. $\endgroup$ – Lord Shark the Unknown Jan 6 '18 at 18:49
  • $\begingroup$ @LordSharktheUnknown: True! I fixed that. Thanks! $\endgroup$ – geguze Jan 6 '18 at 18:52
  • $\begingroup$ Another perspective using representability is $\mathbf{C}(X,\prod_{i\in I}C_i)\cong\prod_{i\in I}\mathbf{C}(X,C_i)$ natural in $X$. The second $\prod_{i\in I}$ is happening in the category of sets. You should work out how this definition is equivalent to Eric Wofsey's and Awodey's definition for binary products in the $I=\{1,2\}$ case. Also, consider the $I=\emptyset$ case. $\endgroup$ – Derek Elkins Jan 6 '18 at 20:07
  • $\begingroup$ I would suggest writing the induced map as $\langle u_i\rangle_{i\in I}$ rather than $\langle u_{i_1},u_{i_2},u_{i_3},\dots,u_{i_j},\dots\rangle$. The latter makes it look like you are assuming $I$ to be countable. $\endgroup$ – Arnaud D. Jan 7 '18 at 11:25
  • $\begingroup$ @ArnaudD.: Good point. Made the change in the text. Thanks! $\endgroup$ – geguze Jan 7 '18 at 18:31
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No, this is not the correct condition. In the universal property of a product, you have just a single object mapping to each of the factors, not different ones. So the correct definition instead would have a single object $X$ with maps $x_i:X\to C_i$ for each $i$, and then there should exist a unique $u:X\to P$ such that $p_i\circ u=x_i$ for all $i$.

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