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I'm confused as to why the coproduct of sets is the disjoint union, and not just the cartesian product (which, I know, is the categorical product here). Having poured over the definition for a while now, I don't see why the cartesian product doesn't equally work. If we'd pick this, then the "inclusion" functions would be standard inclusion.

In the category of modules, for instance, the coproduct is the direct sum. I thought this might suggest that there's problems with infinite products. But, then again, in the formal definition of a disjoint union, the union is taken over an arbitrary set I, so this shouldnt be a problem.

If I were correct, then this would suggest that the cartesian product is isomorphic to the disjoint union (by the universal property, which is obviously not the case (looking at "vectors" in the former, and entries in the latter), so i must be wrong!

I'm a little at a loss, any help would be much appreciated!

Sebastiaan Eeckels
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    What is the "standard inclusion"? Try to write it down and you'll find it doesn't exist. – Qiaochu Yuan Jun 27 '24 at 15:21
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    Further, even in the cases where you can find some "standard inclusion", the cartesian product will usually be too large for the uniqueness part of the universal property to hold. – Rhys Steele Jun 27 '24 at 15:25
  • Thank you Qiaochu Yuan! You're right, I assumed we could just add 0s to each coordinate not used but those ofc don't exist. Also, thank you Rhys Steele! You're also right! I realized you can permute the coordinates around before projecting, resulting in way too many maps. – Sebastiaan Eeckels Jun 29 '24 at 09:01

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Let $0$ denote the initial object, and let $1$ denote the terminal object of some category. Let $\times$ and $\uplus$ denote the product and coproduct, respectively. Recall that in the category of sets, the initial object is the empty set and a terminal object $1$ is a singleton set, and these are not isomorphic.

However, binary products and coproducts cannot coincide in a category where initial and terminal objects are not isomorphic (they are isomorphic e.g. in categories of modules, since $\{0\}$ is both initial and terminal).

Why?

By the universal property of initial and terminal objects, for each object $A$ we have a unique arrow $i_A : 0 \rightarrow A$ and $t_A: A \rightarrow 1$. It follows that the only arrow $0 \rightarrow 0$ is the identity $\mathrm{id}_0$, and the only arrow $1 \rightarrow 1$ is $\mathrm{id}_1$.

Is there a morphism of signature $k: 1 \rightarrow 0$? If we can find such an arrow, then $i_1 \circ k: 1\rightarrow 1$, and by uniqueness $i_1 \circ k = \mathrm{id}_1$. Similarly, $k \circ i_1 : 0 \rightarrow 0$, so $k\circ i_1 = \mathrm{id}_0$. This means that $k$ constitutes an isomorphism between $0$ and $1$. There is, of course, no morphism of signature $k: 1 \rightarrow 0$ in the category of sets.

Now consider the product $0 \times 1$. By definition, we have a projection morphism $\pi_0: 0 \times 1 \rightarrow 0$.

Is there a morphism of signature $m: 1 \rightarrow 0\times 1$? If there is, then we have $\pi_0 \circ m: 1 \rightarrow 0$, so by the previous argument the initial and terminal objects are isomorphic. Needless to say, there can be no morphism $m:1 \rightarrow 0\times 1$ in the category of sets (in concrete terms, the product of an empty set with a singleton set is empty).

However, for the coproduct, there is a morphism of signature $\iota_1: 1 \rightarrow 0 \uplus 1$ by definition. If the initial and terminal objects are not isomorphic, then there is no morphism of signature $1 \rightarrow 0 \times 1$, and so $0 \times 1$ is not the coproduct. In particular, $0 \uplus 1$ does not coincide with $0 \times 1$ in the category of sets.

This is why Qiaochu Yuan told you to try and write down the "standard inclusion" to find out that it doesn't exist. In particular, there is no inclusion of $1$ to $0 \times 1$.

Z. A. K.
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  • The argument above shows that product and coproduct don't always coincide in Set. But keep in mind that in some specific cases, binary products and coproducts may coincide: for example, if $A = {0,1}$, then $A \times A$ is isomorphic to $A \uplus A$. – Z. A. K. Jun 27 '24 at 17:53