What is the simplest way to check whether a given function of two arguments (Its arguments and the value are morphisms of some category.) is a direct product in categorical sense?
How to check whether it is a direct product?
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0What does it mean for a single morphism to be a direct product? – 2012-01-26
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0@Zev Chonoles: From http://en.wikipedia.org/wiki/Product_%28category_theory%29 : "there exists a unique morphism f : Y \to X such that the following diagram commutes ... The unique morphism f is called the product of morphisms" – 2012-01-26
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0I would call the corner $X_1\xleftarrow{\;\;\pi_1\;\;} X_1\times X_2\xrightarrow{\;\;\pi_2\;\;} X_2$ a direct product, not the morphism $f$, but I understand what you're asking now. You should indicate in your question the necessary conditions on the domains and codomains of the morphisms for it to make sense. – 2012-01-26
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0@Zev Chonoles: In fact I have a function which takes two ARBITRARY (with arbitrary domains and codomains) morphisms of certain category. We may restrict this function to take only morphisms with identical domains. I suspect (after this restriction) it will be a direct product in categorical sense. – 2012-01-26
1 Answers
You seem to be asking whether it is possible to give an essentially algebraic axiomatisation of categorical products. The short answer is: yes, but you need some additional data.
Let $\mathcal{C}$ be a category. Suppose we have the following operations:
- For every pair of objects $(A, B)$, another object $A \times B$ and two arrows $\pi_{A,B} : A \times B \to A$, $\pi'_{A,B} : A \times B \to B$.
For every triple of objects $(A, B, C)$ and pair of arrows $f : C \to A$, $g : C \to B$, an arrow $\langle f, g \rangle : C \to A \times B$, such that the following axioms hold:
$\pi_{A,B} \circ \langle f, g \rangle = f$
$\pi'_{A,B} \circ \langle f, g \rangle = g$
For all $h : C \to A \times B$, $\langle \pi_{A,B} \circ h, \pi'_{A,B} \circ h \rangle = h$
Exercise. Verify that the triple $(A \times B, \pi_{A, B}, \pi'_{A, B})$ has the universal property of the product of $A$ and $B$.
Some other universal constructions in categories can also be made essentially algebraic: this is done in the first chapter of Lambek and Scott's Introduction to higher order categorical logic, for example.
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0P.S. If $\mathcal{C}$ is a _small_ category with all binary products, then it can be endowed with these operations. Otherwise one needs to assume a sufficiently large axiom of choice. – 2012-01-26
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01. It seems that you've made a mistake. As far as I understand, it should be $\pi_{A,B} \circ \langle f, g \rangle = f$ not $\pi_{A,B} \circ f = f$. 2. Could you show how uniqueness follows? – 2012-07-22
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0Uniqueness follows from the fact that the pairing is invertible. – 2012-07-23
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0OK, I read through your answer and confirmed it is equivalent to the standard definition. I wonder why your simplified description is missing in textbooks. – 2012-07-23
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0I want to refer to your definition of direct product (not only finite but also infinite) in my research article. What should I write in my bibliography? – 2012-07-23
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0This can be found in Lambek and Scott's _Introduction to higher order categorical logic_, for example. The reason this definition is not often given is because it is _not_ equivalent to the standard definition: in order to get a map $(A, B) \mapsto A \times B$, one must usually invoke a sufficiently powerful form of the axiom of choice. – 2012-07-23
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0I completely don't understand how this is related with axiom of choice. I wrote above that I checked it and found it equivalent to the standard definition. Do I err? – 2012-07-23
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0Zhen Lin, maybe you've overlooked my question? I repeat it again: Why they are not equivalent and how this relates with axiom of choice? Also: It is worth to pay $50 for "Introduction to higher order categorical logic"? (provided that I'll newer read this book from cover to cover but need just one or two paragraphs from it) – 2012-07-23
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0I already explained why the axiom of choice is required. Please reread by comments. – 2012-07-24
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0You have only said: "in order to get a map $(A, B) \mapsto A \times B$, one must usually invoke a sufficiently powerful form of the axiom of choice". It seems to not be required in the case of finite products. Do you mean infinite product? Even in the infinite case, I don't understand why it's required. – 2012-07-24
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0Read _carefully_. I am talking about the assignment of a product $A \times B$ to each pair $(A, B)$, not the existence thereof, or its properties. – 2012-07-24
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0It seems now I understand. A possible reason I was not understanding is that somewhere I've seen a definition of "category with finite products" which includes a definite function which is called "direct product". – 2012-07-24
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0If you are given the operation (A,B)↦A×B, then a product preserving functor between categories with finite products must preserve the specified product. If you require only that there be an object A times B with the required properties, then a product preserving functor must take a product to SOME product of the images of A and B (of course, they are all naturally isomorphic). Note: This is a fine point that hardly ever makes any difference in practice! See Toposes, Triples and Theories at www.tac.mta.ca/tac/reprints/articles/12/tr12.pdf, page 141. – 2012-07-24