Cardinality of the product category

Question

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.

Answer 1

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

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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.

Answer 2

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).