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After reading this post from Milewski's blog, I think the definition of product is defined as follows

Given A and B are objects in a Category. $X$ is a product of A and B if and only if there is a pair of morphisms:

$$\pi_1 :: X \rightarrow A, \quad\pi_2 :: X \rightarrow B$$ and for any "other" object $Y$ with a pair of morphisms $$q_1 :: Y \rightarrow A, \quad p_2 :: Y \rightarrow B$$

There is a unique morphism $h$, such that $q_1 = \pi_1 \circ h$ and $q_2 = \pi_2 \circ h$

I highlight "other" in the definition as I also see this in other places 2. If we have "other" in the definition can we still prove that all the products of $A$ and $B$ satisfying the definition are isomorphic to each other?

My attempt to prove it is to show that if $X$ and $X'$ both satisfying the definition of the product $A \times B$, then we know there is a morphism $h: X \rightarrow X'$ and a morphism $h': X' \rightarrow X$. I want to show that $h' \circ h = id_X$ by choosing $Y=X$ and show that $id_{X}$ is the unique morphism from $X$ to $X$, but it seems that the "other" requirement in the definition prevents me from doing that.

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    $\begingroup$ You proof is more than fine; the wording in the definition could be improved: the 'for any other' does not imply distinct. $\endgroup$
    – qualcuno
    Mar 4, 2021 at 23:31
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    $\begingroup$ Your proof is right. And your comment about the phrasing of the definitions is quite right! Pace @guidoar the word "other" does imply distinct and should have been omitted from the definition. Another very common blooper along these lines is to say, "let $x, y$ be two elements of $X$, then ...": which excludes the possibility that $x = y$ (because, for example, $1$ and $1$ are not two elements of the integers). $\endgroup$
    – Rob Arthan
    Mar 5, 2021 at 0:01
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    $\begingroup$ @RobArthan my wording was rather unfortunate. What I intended to say is, indeed, that 'other' should be ommited because it conveys something that is not requiered in the definition of product. Thanks for pointing that out :) $\endgroup$
    – qualcuno
    Mar 5, 2021 at 0:05
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    $\begingroup$ @guidoar no worries. I love these cases where we can improve on the precision of our mathematical use of language just by crossing out noise words. $\endgroup$
    – Rob Arthan
    Mar 5, 2021 at 0:10
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    $\begingroup$ Although colloquially “other” implies distinct from the one, in mathematics it need not (just like colloquially “or” is exclusive, but in mathematics it almost never is). When one truly wants to say that there is something else distinct from the one, we tend so say so explicitly rather than implicitly by the choice of connectives. The word “distinct” tends to show up explicitly. That said, there is no reason to have a potentially confusing definition when one can give a precise one. But if this was stated colloquially rather than formally, that would explain it. $\endgroup$ Mar 5, 2021 at 0:10

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As discussed in the comments, the term "other" here is not intended to require $Y$ to be distinct from $X$. So, there is no difficulty carrying out your argument. This usage of "other" is fairly common in mathematics, though I would personally consider it ambiguous enough to recommend avoiding it in formal mathematical writing.

If you interpret the definition to require that $Y\neq X$ in order to use the universal property, then a product need not be unique up to isomorphism. For example, consider the following subcategory of the category of sets. The objects are $A=\{0\},B=\{1\},X=\{2,3\},$ and $X'=\{4,5\}$. The morphisms are the identity functions, the unique functions from each of $X$ and $X'$ to $A$ and $B$, the constant functions $X\to X$ and $X'\to X$ with constant value $2$, and the constant function $X'\to X'$ and $X\to X$ with constant value $4$. It is then straightforward to verify that both $X$ and $X'$ are "products" of $A$ and $B$ in this category if you require $Y\neq X$ in the universal property, but are not isomorphic. For instance, to check that $X$ is a "product" of $A$ and $B$, the only choice of $Y$ you have to consider is $X'$, and the unique map $X\to X'$ (the constant $4$ map) has the required properties.

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