1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What does it mean for a single morphism to be a direct product? $\endgroup$ – Zev Chonoles Jan 26 '12 at 16:25
  • $\begingroup$ @Zev Chonoles: From 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" $\endgroup$ – porton Jan 26 '12 at 16:31
  • $\begingroup$ I 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. $\endgroup$ – Zev Chonoles Jan 26 '12 at 16:36
  • $\begingroup$ @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. $\endgroup$ – porton Jan 26 '12 at 16:50
2
$\begingroup$

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:

    1. $\pi_{A,B} \circ \langle f, g \rangle = f$

    2. $\pi'_{A,B} \circ \langle f, g \rangle = g$

    3. 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.

$\endgroup$
  • $\begingroup$ P.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. $\endgroup$ – Zhen Lin Jan 26 '12 at 17:49
  • $\begingroup$ 1. 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? $\endgroup$ – porton Jul 22 '12 at 18:39
  • 1
    $\begingroup$ Uniqueness follows from the fact that the pairing is invertible. $\endgroup$ – Zhen Lin Jul 23 '12 at 2:15
  • 1
    $\begingroup$ OK, I read through your answer and confirmed it is equivalent to the standard definition. I wonder why your simplified description is missing in textbooks. $\endgroup$ – porton Jul 23 '12 at 8:29
  • 1
    $\begingroup$ You 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. $\endgroup$ – porton Jul 24 '12 at 7:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.