# Products and coproducts in $\mathsf{P}$ (the power set considered as a category)

I was advised to explore products and coproducts in $$\mathsf{P}$$ in order to improve my understanding of products and coproducts in general. Here I will type up my understanding.

Power set as category

Given set $$X$$ (which I suppose we should require to be non-empty), $$(\mathcal P(X), \subseteq)$$ is a partially ordered set. We can consider the partially ordered set to be a category wherein the objects are elements of $$\mathcal P(X)$$, and there is a morphism $$f \colon A \to B$$ for $$A, B \in \mathcal P(X)$$ if and only if $$A \subseteq B$$. This means that any hom set contains either zero or one morphism. The composition rule is satisfied because of transitivity of the order relation.

Composition of morphisms is associative because if $$h \colon X \to Y, g \colon Y \to Z$$, and $$f \colon Z \to W$$, then $$f(gh) = (fg)h$$ because both represent the true statement that $$X \subseteq W$$. Identity morphisms exist because of the reflexivity of the order operation.

The product in $$\mathsf{P}$$

Given category $$\mathsf C$$, index set $$\mathsf I$$ and indexed objects $$X_i$$ from the category, the product of $$\{ X_i ; i \in \mathsf I \}$$ is an object of $$\mathsf C$$, denoted $$\prod X_i$$, together with an indexed family of morphisms $$\pi_i \colon \prod X_i \to X_i$$ such that for any object $$Y$$ in $$\mathsf C$$ and indexed family of morphisms $$f_i \colon Y \to X_i$$, there is a unique map $$f \colon Y \to \prod X_i$$ such that $$\pi_i f = f_i$$ for all $$i \in \mathsf I$$.

If we translate this into the more intuitive language of $$\mathsf{P}$$, then $$\prod X_i$$ is an object such that for any object $$Y$$ such that $$Y \subseteq X_i$$ for all $$i \in \mathsf I$$, we have $$Y \subseteq \prod X_i \subseteq X_i$$. The natural choice is $$\prod X_i := \inf \{X_i\}$$, because if we choose anything smaller then we might contradict our requirement that $$Y \subseteq \prod X_i$$. Uniqueness is not an issue because there is only one morphism from $$Y$$ to $$\prod X_i$$.

The coproduct in $$\mathsf{P}$$

Since the coproduct is dual to the product, I should be able to reverse arrows in the preceding abstract description in order to get the correct abstract description of a coproduct. Therefore, the coproduct of $$\{ X_i ; i \in \mathsf I \}$$ is an object of $$\mathsf C$$, denoted $$\coprod X_i$$, together with an indexed family of morphisms $$\pi_i \colon X_i \to \coprod X_i$$ such that for any object $$Y$$ in $$\mathsf C$$ and indexed family of morphisms $$f_i \colon X_i \to Y$$, there is a unique map $$f \colon \coprod X_i \to Y$$ such that $$f\pi_i = f_i$$ for all $$i \in \mathsf I$$.

Translating into the language of $$\mathsf{P}$$ again, $$\coprod X_i$$ is an object such that for any object $$Y$$ such that $$X_i \subseteq Y$$ for all $$i \in \mathsf I$$, we have $$X_i \subseteq \coprod X_i \subseteq Y$$. The natural choice is $$\coprod X_i := \sup \{ X_i ; i \in \mathsf I \}$$ because if we choose anything larger we might contradict our requirement that $$\coprod X_i \subseteq Y$$. Uniqueness is not an issue because there is only one morphism from $$\coprod X_i$$ to $$Y$$.

Questions:

Is my work above correct? (It's my understanding that $$\pi$$ is not used for the coproduct but I didn't feel like changing it since I'm emphasizing that the coproduct at least in the abstract is the reverse of the product.)

I'm under the impression that universal properties are worth learning about. How should I go about learning about universal properties in relation to what I'm doing here? Should I just look at examples?

I appreciate any help.

• What is "$\mathsf{Poset}$"? The category of posets or one particular poset, considered as a category? I would assume it means the first but you seem to be answering as though it is the second. Sep 25 '20 at 21:48
• @EricWofsey Looking at my text, I mean one poset considered as a category. Should I modify the title of my post? If so, how? Sep 25 '20 at 21:50
• It is very unusual to use the notation $\mathsf{Poset}$ to refer to such a category. I would suggest you give it a different name and explicitly define it. Sep 25 '20 at 21:51
• Usually for a poset $(P,\leq)$, the partial order $\leq$ is understood and the poset category is denoted with the same letter, but different font. As in ${\sf P}$, with ${\rm Obj}({\sf P}) = P$ and $|{\rm Mor}_{{\sf P}}(a,b) |= 1$ if $a \leq b$, empty else. Sep 25 '20 at 21:57
• Yes, your understanding is correct. More examples are always a good idea. Some other good examples to test your understanding on are: the category of [sets, topological spaces (especially for counter-examples), graphs, groups, vector spaces], and also categories of functors if you are familiar with them. Sep 25 '20 at 23:15

In this particular example they're specifically the intersection and the union of the given subsets: $$\prod_iX_i=\bigcap_i X_i\,\phantom{.}\\ \coprod_iX_i=\bigcup_i X_i\,.$$
• It's worth pointing out that those equivalences are specific to $(\mathcal{P}(X),\subseteq)$, and might not hold if some subsets or inclusions are missing. Oct 24 '20 at 13:39
• In particular, they generally won't hold for an arbitrary subcategory $D\subseteq (\mathcal{P}(X),\subseteq)$. Oct 24 '20 at 13:46