0
$\begingroup$

Let $\mathcal{C}$ be a category and let $A,B$ be objects of $\mathcal{C}$. Let $\pi_A$ and $\pi_B$ be morphisms in $\mathcal{C}$ such that $\pi_A:A\times B\to A$ and $\pi_B: A\times B\to B$. Now, the universal property says that for each object $X$ in $\mathcal{C}$ and every pair of morphisms $f_A, f_B$ there exists a unique morphism $f: X\to A\times B$.

I'm having a really hard time understanding why this must be true or maybe I'm missing some information or definition.

The universal property on any kind of structure is very loosely defined, that I cannot find a solid definition anywhere on the internet. How can this be possibly true for every $X,f_A$ and $f_B$ here? Are there any restriction on any of the given morphisms that I failed to mention?

Any help would be appreciated.

https://i.imgur.com/AcdaqkI.png

EDIT: I am unable to include the image in this question for whatever reason. see link

$\endgroup$
9
  • $\begingroup$ The statement is that for every pair of morphisms $f_A : X \to A$ and $f_B : X \to B$, there is a unique $f : X \to A \times B$ such that $\pi_A \circ f = f_A$ and $\pi_B \circ f = f_B$. $\endgroup$ Sep 13, 2021 at 21:47
  • 1
    $\begingroup$ Generally, one picks out a certain specific category (such as the category of sets), picks out a choice of $A \times B$, $\pi_A$, and $\pi_B$, and then proves that the given property is true. "why is that true" depends on which category you're talking about. $\endgroup$ Sep 13, 2021 at 21:50
  • 1
    $\begingroup$ "the product of $A$ and $B$ exists" is defined to mean that there exists an object $A \times B$ and arrows $\pi_A, \pi_B$ such that the property holds. $\endgroup$ Sep 13, 2021 at 21:53
  • 1
    $\begingroup$ You're missing the point here. Whenever we have a category $C$ and objects $A$ and $B$ in this category, the statement "the product of $A$ and $B$ exists" means, by definition, that there exists an object $A \times B$ and two arrows $\pi_A : A \times B \to A$, $\pi_B : A \times B \to B$ such that for all objects $X$ and arrows $f_A : X \to A$, $f_B : X \to B$, there is a unique arrow $f : X \to A \times B$ satisfying $f_A = \pi_A \circ f$ and $f_B = \pi_B \circ f$. This is the definition of "the product exists". $\endgroup$ Sep 13, 2021 at 22:00
  • 2
    $\begingroup$ Oh so the "there exists" part is also part of the definition of "the product exists"? That makes much more sense. Thanks man $\endgroup$ Sep 13, 2021 at 22:05

1 Answer 1

1
$\begingroup$

"why is that true ?" : we define the tuple $(P, \pi_A,\pi_B)$ to be a product if it satisfies this property. So it is true for the product because we defined products to be those things that satisfy it.

"How can this possibly true" : think about the category of sets, and the usual cartesian product of sets $A\times B$ with projections $(a,b)\mapsto a, (a,b)\mapsto b$. Can you try to prove and understand why it satisfies this condition ?

Now can you try to show something similar for the cartesian product of groups ? of rings ?

Note that a product in a category $C$ need not have anything to do with the usual cartesian product of sets, even when $C$ is a category of "sets with structure". It is very often true that it is related to the product of sets for categories of sets with algebraic structures, but then that is something you have to prove, and it does not follow from the definition of a product.

$\endgroup$
1
  • $\begingroup$ Thanks! I was able to convince myself this holds after re-learning the definition. Stupid me. $\endgroup$ Sep 13, 2021 at 22:35

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.