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It's clear that $X \times Y$, the Cartesian product of two sets, satisfies the universal property of the product of $X$ and $Y$ in the category $\mathsf{Set}$ of sets. It is possible to show that $Y \times X$ satisfies the same property, so that it is "the" product of $X$ and $Y$ as much as $X \times Y$ is. Moreover, the two are canonically isomorphic, that is, there exists a unique isomorphism between them: this is a consequence of the universality of the product.

My question then is: are there other objects of $\mathsf{Set}$ which are products of $X$ and $Y$?

Considering just equinumerous sets shouldn't work in principle, because they are isomorphic to $X \times Y$ but not uniquely so. Then I don't have a lot of other ideas... I know that $\prod_{i \in I} X^I \cong X^I$, but I can't generalize this to the case where $X \neq Y$.

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Given sets $X$ and $Y$, any set $P$ of cardinality $|X| \cdot |Y|$ can serve as a product of $X$ and $Y$, provided you pick appropriate projections $X \xleftarrow{p_1} P \xrightarrow{p_2} Y$. Remember an object of a category alone is not a product, rather, it is the object together with the projection maps.

Explicitly, let $X \xleftarrow{\pi_1} X \times Y \xrightarrow{\pi_2} Y$ be the usual cartesian product and projection functions. Since $|X \times Y| = |P| = |X| \cdot |Y|$, there is a bijection $f : P \to X \times Y$. Now define $p_1 = \pi_1 \circ f$ and $p_2 = \pi_2 \circ f$. You can verify that $(P,p_1,p_2)$ satisfy the appropriate universal property.

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  • $\begingroup$ Accepting this as more concise. Thank you Clive! $\endgroup$ – mattecapu Apr 7 '18 at 20:31
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    $\begingroup$ @mattecapu Just to say it outright, as long as neither $X$ nor $Y$ is empty, this means that there is actually a proper class of sets that are products of $X$ and $Y$. $\endgroup$ – Derek Elkins Apr 8 '18 at 0:13
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You have an important misunderstanding of what a product is. A product of $X$ and $Y$ is not an object $X\times Y$. It is a triple $(X\times Y, p,q)$, where $X\times Y$ is an object and $p:X\times Y\to X$ and $q:X\times Y\to Y$. It is common to abuse notation and refer to $X\times Y$ alone as the product, but the maps are crucial (just like we frequently refer to a group by just its underlying set when actually the group also includes the operation).

So, when we say a product is unique up to unique isomorphism, we are talking about isomorphisms of triples, not just of objects. In other words, if $(Z,p,q)$ and $(Z',p',q')$ are both products of $X$ and $Y$, there is a unique $f:Z\to Z'$ which is not just an isomorphism but also satisfies $p'f=p$ and $q'f=q$.

Moreover, given a product $(Z,p,q)$ of $X$ and $Y$, an object $Z'$, and an isomorphism $f:Z\to Z'$, we can define $p'=pf^{-1}$ and $q'=qf^{-1}$, and it is easy to see that $(Z',p',q')$ is also a product of $X$ and $Y$ (with $f$ being the unique isomorphism between the triples $(Z,p,q)$ and $(Z',p',q')$).

So, in the case of sets, any set $Z'$ equinumerous with $X\times Y$ can be given the structure of a product of $X$ and $Y$, by choosing some bijection $f:X\times Y\to Z'$. Such a set $Z'$ is isomorphic to $X\times Y$ in many different ways, but in only one way once you fix product structures on both objects.

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    $\begingroup$ Yes, I was aware that products are actually things like $X \leftarrow X \times Y \to Y$, but it didn't occur to me the unique isomorphism was actually about this triple and not just the product object. Thank you! $\endgroup$ – mattecapu Apr 7 '18 at 20:29

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