Doubt trying to understand the product in category theory?

Question

I'm reading Walters' Categories and Computer Science, here we have the definition of product:

I made a Mathematica code just to "see the definition working" but am confused with the following issue:

• I defined my objects as elements of $A=\{\{\},\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$.
• I defined my arrows as $a\to b$ if $a \subset b$.
• I want to find the product of $X=\{1, 2\}$ and $Y=\{1, 3\}$, looking at the definition, I think I need to collect all $i\in A$ such that we have $X\times Y \to X$ and $X\times Y \to Y$, there are two possibilities: $\{\},\{1\}$.
• Now I think that we need to check if for all $Z\in A$, we have the arrows: $Z\to X\times Y, Z \to X, Z \to Y$ for $X\times Y=\{\}$ and then for $X\times Y=\{1\}$.

The trouble I am facing is the following: The code I made takes each possibility for $X\times Y$ and highlights the following edges:

• $X\times Y \to X$
• $X\times Y \to Y$
• $Z\to X\times Y$
• $Z \to X$
• $Z \to Y$

For all $Z\in A$ if all them exist for some $Z$. It returns me the following graphs:

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Question: I am a bit confused, what is the actual product here? Is it $\{\}$?Is it $\{1\}$? Is it both? I may be confusing something really silly but I am really lost here. I suspect $\{\}$ is not a product here but I don't know why.

Answer 1

In your case the product corresponds to the intersection.

Things you have:

1. $p_1 : \{1\}\rightarrow \{1,2\}$
2. $p_2 : \{1\}\rightarrow \{1,3\}$
3. $q_1 : \{\} \rightarrow \{1,2\}$
4. $q_2 : \{\} \rightarrow \{1,3\}$

Two candidates for a product with their respective projections. However $\{1\}$ has something stronger, and it's that for $\{\}$ and its projections $q_1$ and $q_2$, there is a unique map $\{\}\rightarrow\{1\}$ call it $f$ such that $q_1 = p_1\circ f$ and $q_2 = p_2\circ f$.

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I think of it this way:

We say an object is a "candidate for a product" if it has the projections. (Formally we call it a "cone".)

Then, a product is a candidate for a product with "maximal information".

As you can see, we can "recover" $q_1$ from $p_1$ and $q_2$ from $p_2$ assuming $f$. So in a sense $p_1$ and $p_2$ "contain more information". And that's what's so fundamental about this definition of product.

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Importantly, you don't have a map from $\{1\}$ to $\{\}$ that has the same effect. Else it could happen that both were products.

Answer 2

Now I think that we need to check if for all $Z\in A$, we have the arrows: $Z\to X\times Y, Z \to X, Z \to Y$ for $X\times Y=\{\}$ and then for $X\times Y=\{1\}$.

I think this is where your confusion lies, since you don't have these arrows in logical order. We don't need universally arrows $Z \to X \times Y$, $Z \to X$, and $Z \to Y$. Instead, we have an arrow $Z \to X \times Y$ whenever there are arrows $Z \to X$, and $Z \to Y$ (subject to a uniqueness condition).

As you correctly surmise, $X \times Y$ in your example can only be either $\{\}$ or $\{1\}$ because those are the only objects with arrows to both $\{1, 2\}$ and $\{1, 3\}$. Let's prove that $\{\}$ is not the product.

For $\{\}$ to be the product, we need for every object $Z$ and every pair of arrows $f : Z \to X$ and $g : Z \to Y$, an arrow $\langle f, g \rangle : Z \to \{\}$. As pointed out before, if we have these arrows $f$ and $g$, $Z$ must be either $\{\}$ or $\{1\}$. Let's check with $Z = \{1\}$. $f$ and $g$ can only be one thing, so let's check that there is an arrow $Z \to \{\}$. But since $Z$ is $\{1\}$ and there is no arrow $\{1\} \to \{\}$, $\langle f, g \rangle$ can't exist. This proves that $\{\}$ cannot be the product.

Spoilers below

You might try proving that in general, the product of two sets in your category are their intersection. Similarly, the coproduct is the union.