Universal Property and Fibered Product

Question

I'm working through Vakil's excellent The Rising Sea notes, and in an exercise, the following question is posed:

If $X$ is a topological space, show that the fibered product always exists in the category of open sets of $X$, by describing what a fibered product is.

Now I know intuitively the fibered product of 2 open sets is nothing but the intersection as the maps between them are that of inclusion. But I'm having trouble proving that explicitly using the universal property. Can anybody help me out?

Answer 1

So lets take two open sets $U,V \subseteq Y \subseteq X$, where $Y$ is open as well.

That means we have two morphisms $U \to Y$, $V \to Y$. Now the fibered product of those morphisms (lets call it $W$) is an object of the same category - hence it is an open subset of $X$. Furthermore we need some maps $W \to U$, $W \to V$. By the definition of our category this translates into: $W \subseteq U$, $W\subseteq V$. Hence we have $W \subseteq U\cap V$ for sure.

Now we want the following: Given any open subset $W' \subseteq X$ together with morphisms $W' \to U$, $W'\to V$ that make the appropriate diagram commute (which is trivial in our case, do you see why?), we want a morphism $W' \to W$ that makes all "new" diagrams commute (again, it is trivial that all diagrams commute once we have a morphism - that is all we need to check).

$W' \to W$ translates to $W' \subseteq W$. Hence intuitively we'd expect $W$ to be the biggest open set contained in both $U, V$. As you've already said, that is $U \cap V$. Now why does $U \cap V$ satisfy the universal property of the fiber product? Because if $W' \subseteq U$, $W' \subseteq V$ it follows that $W' \subseteq U \cap V$, which exactly means that there is a morphism $W' \to U\cap V$. Again, by the very definition of our category it is clear, that this morphism is unique - there is at most one morphism between any two objects.

Answer 2

The category of open sets is a poset; that is there's at most one morphism between objects; here there's a morphism $U\to V$ iff $U$ is an open subset of the open subset $V$. In general if one has arrows $U\to W$ and $V\to W$ in a poset, then they have a pullback (fibre product) iff $U$ and $V$ have a least upper bound $Y$ in the poset. Then $Y\to U$ and $Y\to V$ finish off the pullback square.

In the category of open sets $U\cap V$ is an infimum of $U$ and $V$; it is an open subset of $U$ and $V$, and each open subset of $U$ and $V$ is an open subset of $U\cap V$.

Answer 3

influenced by Angina Seng's answer:

In this category of open subset of $X$.

$U,V\subset X$, $$U\to V\quad unique\quad \Leftrightarrow\quad U\subset V$$

Claim: $U\times_WV=U\cap V$ It is indeed a cone (satisfying the commutative square). Now given any $P\subset X$ with $P\to U$ and $P\to V$ it means $P\subset U\cap V$ then $P\to U\cap V$ and it is unique proving the limit of the diagram(or fibered product in this case).