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Under which additional conditions $a\times b = c\times d \Rightarrow a=c\wedge b=d$ (where $\times$ is a categorical product)?

For example, in the case of Cartesian product, for this is enough when the factors are non-empty. Can this be generalized?

I'm also interested about the similar construction with infinite product.

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    Am I misreading this? Since $a\times (b\times c)=(a\times b)\times c$, but $a\neq a\times b$, you don't even get this for Cartesian products. Unless you mean absolute equality with "=", but products are not defined in Category theory with absolute equality2012-07-05
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    @ThomasAndrews: I am similarly puzzled by the question, but I don't think your example works. $a\times(b\times c)$ has elements of the form $\langle a,\langle b,c\rangle\rangle$, where $\langle a,b\rangle$ means something like $\{\{a\}, \{a, b\}\}$, but $(a\times b)\times c$ has elements of the form $\langle\langle a,b\rangle, c\rangle$. As in category theory, the sets are isomorphic but unequal.2012-07-05
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    @ThomasAndrews: I don't know where you've got $a\times (b\times c)$ from, i ask only about binary product (not counting my question below about infinite case). OK, we should replace equality with isomorphism. Maybe you've misunderstood what ∧ means? It is just conjunction.2012-07-05
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    @MarkDominus Which is why I asked about absolute equality, and not category isomorphism.2012-07-05
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    $a\times(b\times c)$ is a binary product of $a$ with $b\times c$2012-07-05
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    @porton If you take = to mean categorical isomorphism, then the implication does not hold for sets, since in the category of sets $a\times b $ is isomorphic to $b\times a$.2012-07-05
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    Guys, you are right, my question is mis-formulated, because it may be about only in precision of an isomorphism. You may close my question.2012-07-05
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    I think there is an interesting question hiding in here, but I'm not sure what it is.2012-07-05
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    Just because it is misformulated, doesn't mean it should be closed: the problems are mathematical problems, after all. Someone should write up the problems in the comments in an answer.2012-07-05
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    I bet every Cartesian category is equivalent to a category where the proposition is true. (with = meaning equality, and x being a bifunctor)2012-07-06
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    @Hurkyl is the one getting closer to the heart of the question.2012-07-06
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    @Hurkyl: If by cartesian category you mean a category with all finite products, then yes. Fix a choice of product bifunctor, and define a new category whose objects are finite strings of objects of the original category and morphisms are the obvious ones. Note that this new category also has a strictly associative product.2012-07-07
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    @ZhenLin: A category with a strictly associative product must fail the condition that $a \times b = c \times d \implies a = c \wedge b = d$.2012-07-07
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    Oops – yes. But that's still easy to fix. Take the category whose objects are finite binary trees whose leaves are objects of the original category.2012-07-07

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