Let $X\times Y$ be the product of objects $X$ and $Y$ in a category $C$, defined by the universal property of products. Is it true that the existence of the product $X\times (Y\times Z)$ implies the existence of $X\times Y$?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
No. Consider the partial order $\mathbb{N}\cup\{\infty_1, \infty_2\}$ where $\infty_1$ and $\infty_2$ are incomparable to each other and greater than all natural numbers. Taken to be a category in the usual way (a unique morphism $x \to y$ if $x \le y$), products here are greatest lower bounds.
Then if $X = \infty_1$ and $Y = \infty_2$, the greatest lower bound of $X$ and $Y$ doesn't exist (the lower bounds are all natural numbers). However, if we take $Z = 0$, then $Y \times Z$ exists and equals $0$. Similarly, $X \times (Y \times Z)$ exists and equals $0$.
$\begingroup$
$\endgroup$
3
No. Consider the following minimal example. Consider the category with objects
$$ A, X, Y, Z, Y \times Z, X \times (Y \times Z)$$
and the morphisms
$$\begin{align} \pi_1: X \times (Y \times Z) &\rightarrow X \\ \pi_2: X \times (Y \times Z) &\rightarrow Y \times X \\ \pi'_1: Y \times Z &\rightarrow Y \\ \pi'_2: Y \times Z &\rightarrow Z \\ a_1:A &\rightarrow X \\ a_2: A &\rightarrow Y \end{align}$$
and their compositions i.e. $\pi'_1 \circ \pi_2$ and $\pi'_2 \circ \pi_2$.
answered Mar 20, 2020 at 9:01
-
$\begingroup$ In this example, $X\times (Y\times Z) $ is the product of $X $ and $Y $. But you can fix it by adding another object with morphisms to both $X $ and $Y$. $\endgroup$ Mar 20, 2020 at 9:09
-
$\begingroup$ Jeremy Rickard, that's what I just thought of... $\endgroup$ Mar 20, 2020 at 9:13
-