In some old notes I found the following formula for the cardinality of the cardinality of objects and arrows in a product category:

$$ |Ob(C \times D)| = |Ob(C)| \cdot |Ob(D)| \\ |Hom(C \times D)| = |Ob(C)| \cdot |Hom(D)| + |Ob(D)| \cdot |Hom(C)| $$

I also noted that the formula for arrows excludes identities.

While the formula for the formula for objects seems obviously true to me, I can't really confirm the second formula. E.g. when I write down the product category of the free quiver, I only get 12 arrows.

Is the formula wrong? I tried searching for it, but I didn't find it again.

  • 1
    $\begingroup$ The morphisms in a category C, or D (I guess). $Hom$ doesn't refer to Hom functor in this case, but to the collection of arrows in a category, just to be clear. $\endgroup$ – hgiesel Jan 4 '18 at 16:04

An arrow in the product category is simply given by a pair of arrows from your categories $C$ and $D$, so $Hom(C\times D)$ should simply be $Hom(C)\times Hom(D)$, which would give you $$|Hom(C\times D)|=|Hom(C)|\cdot |Hom(D)|$$ However, you mention that the formulat excludes the identities (for some reason). If we write $Hom'(C)$ to be the collection of non-identities arrow in $C$, then identifying every object with its identity gives us a bijection $Hom(C)\cong Hom'(C)\sqcup Ob(C)$ and thus we have $$|Hom(C)|= |Hom'(C)|+| Ob(C)|.$$ In particular, \begin{align}|Hom(C\times D)| = & |Hom(C)|\cdot |Hom(D)| \\ = & (|Hom'(C)|+| Ob(C)|)\cdot (|Hom'(D)|+| Ob(D)|)\\ = & |Hom'(C)|\cdot |Hom'(D)|+|Hom'(C)|\cdot |Ob(D)|\\ & +|Ob(C)|\cdot |Hom'(D)|+|Ob(C)|\cdot |Ob(D)|;\end{align} and thus $$|Hom'(C\times D)|=|Hom'(C)|\cdot |Hom'(D)|+|Hom'(C)|\cdot |Ob(D)| +|Ob(C)|\cdot |Hom'(D)|.$$

I think your $Hom$ is what I denoted $Hom'$, and your second formula is missing a term; in the example of the free quiver with itself, you would get that the total number of non-identity arrows on the product is $2\cdot 2+2\cdot 2+2\cdot 2=12$, since there are two objects and two non-identity arrows.


The formula seems to work fine in this case. Arrows in the product category are pairs of arrows, each arrow in the corresponding category.

The free quiver has two objects $E$ and $V$, two arrows from $E$ to $V$ and the identities on them.

The product category of this category with itself has $4$ objects (each with one identity), $4$ maps from $E\times E$ to $V\times V$, $2$ maps from $E\times E$ to $E\times V$ and another $2$ from $E\times E$ to $V\times E$, and finally $2$ maps from $E\times V$ to $V\times V$ and another $2$ from $V\times E$ to $V\times V$.

The total is $16$, which agrees with your formula (unless I did something wrong).

  • $\begingroup$ Ok, you included the identities in final amount of arrows. But just think about what happens, when I remove, e.g. $t$ from the category, now there are 5 arrows + 4 identities = 9 arrows, however it should be only 4 (2 * 1 + 2 * 1), which would be only the identities. I'm not really convinced yet... $\endgroup$ – hgiesel Jan 4 '18 at 16:16
  • $\begingroup$ I am sorry, I didn't read that part of the question. What is the precise statement of "excludes identities"? Do you mean that you don't take the identities into account for the cardinality of the family of arrows? In that case, the formula doesn't seem to work. On the RHS you would have (for the free quiver example) 8 and on the LHS 12. Right? $\endgroup$ – Pedro Jan 4 '18 at 16:42
  • $\begingroup$ As I have said, I'm unsure of the formula myself. If I understand my own writings correctly, then yes, there should 16 arrows, excluding identities. $\endgroup$ – hgiesel Jan 4 '18 at 16:44
  • $\begingroup$ I think the fact that the total agrees with the formula is really a coincidence here. $\endgroup$ – Arnaud D. Jan 4 '18 at 16:49
  • $\begingroup$ @ArnaudD. I agree, I was thinking about it and I couldn't see why that expression would come out at all. Thank you for your answer! $\endgroup$ – Pedro Jan 4 '18 at 16:54

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.