Cartesian products in categories of subobjects

Question

Let $\mathcal{C}$ be a category and $A$ be an object of $\mathcal{C}$. If the inclusion functor from the category $Sub_{\mathcal{C}}(A)$ of subobjects of $A$ (objects are monomorphisms of $\mathcal{C}$ into $A$) into the slice category $\mathcal{C} / A$ has a left adjoint, then (binary) Cartesian products in $Sub_{\mathcal{C}}(A)$ are pullbacks in $\mathcal{C}$. Are there examples of categories $\mathcal{C}$ with some object $A$ such that (binary) Cartesian products in $Sub_{\mathcal{C}}(A)$ are not pullbacks in $\mathcal{C}$?

Answer 1

We can construct a formal, almost* universal counterexample as follows.

Consider the category generated by the following commutative diagram.

Commutativity involves the square in the bottom right and four triangles in the top left, and of course everything what follows from it. The two morphisms $U \cap V \rightrightarrows T$ are not equal, though.

In this category, you can prove that

• $\mathrm{Sub}(A)$ has exactly four objects: $A \xrightarrow{\mathrm{id}} A$, $U \to A$, $V \to A$ and $U \cap V \to A$.
• $U \cap V \to A$ is the product of $U \to A$ and $V \to A$ in the category $\mathrm{Sub}(A)$
• $T \to U$ and $T \to V$ are no monomorphisms (this is what the two morphisms $U \cap V \rightrightarrows T$ are for*)
• There is no morphism $T \to U \cap V$. In particular, $U \cap V$ cannot be the pullback of $U \to A$ and $V \to A$.

*First, I thought about introducing another object $B$ with two morphisms $B \rightrightarrows T$ that are equalized by $T \to U$ and $T\to V$, to witness that they are no monomorphisms. But since then no non-identity morphism has codomain $B$, the $B \rightrightarrows A$ will be monomorphisms and we need a morphism $B \to U \cap V$ to ensure that $U \cap V$ is the product. I think we can simplify this by just letting $B = U \cap V$. This makes the example non-universal, but a bit simpler to work with.