# Direct product in Set with emptyset

I've (partially) read some books about category theory. But only now attempting to put it into research practice I noticed that I do not really understand direct products.

Consider a product $A\times B$ of arrows $x_1: X\rightarrow A$ and $x_2: X\rightarrow B$.

Let now the category Set and $A=\varnothing$. Then there are no arrow $x_1: X\rightarrow A$.

So direct product (in Set) with an empty set does not exist. (I previously though that it exist. Was I wrong?)

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I assume you're writing $X$ for $A \times B$. The conclusion has to be that $\emptyset \times B = \emptyset$. Check that this satisfies the universal property! – Dylan Moreland Jul 22 '12 at 19:01
@DylanMoreland: No, $X$ is an arbitrary set. – porton Jul 22 '12 at 19:03
There is an arrow $X \to \emptyset$ if $X = \emptyset$ ! – Ted Jul 22 '12 at 19:04
@porton I don't see how that shows that $A \times B$ does not exist. The definition of the product of two elements in a category does not say that for any $X$ in the category there exists an arrow $X \to A$. And with $A = \emptyset$ there is only one such $X$ and one such arrow. – Dylan Moreland Jul 22 '12 at 19:05

To check that $\emptyset$ satisfies the categorical definition of the product $M\times\emptyset$, we need to specify morphisms $\pi_1:\emptyset\to M$ and $\pi_2:\emptyset\to\emptyset$ and then check the universal property. The morphisms are easy, since there is only a single map from $\emptyset$ to any set, namely the empty map. Now given any set $Y$ and morphisms $f_1:Y\to M$ and $f_2:Y\to\emptyset$, there must be a unique $f:Y\to\emptyset$ such that $\pi_1\circ f=f_1$ and $\pi_2\circ f=f_2$. Note that $Y$ must be empty, since otherwise there are no such morphisms $f_2$. Since $Y$ is empty, $f_1$ and $f_2$ are empty maps, and there is indeed a unique $f:Y\to\emptyset$ with the required property, namely the empty map. Thus the universal property is fulfilled, and $\emptyset$ is indeed the categorical product $M\times\emptyset$.