Can a subcategory of a category which does not have products have products? The title isn't quite what I want to ask. I want to ask for something stronger: Is there a category $C$ and a full subcategory $C_0$ of $C$ with objects $A, B\in C_0$ such that there is no product in $C$ of $A$ and $B$ but

*

*$C_0$ has all products (i.e., the diagonal functor $C_0\to C_0^{\{\bullet\bullet\}}$ has a right adjoint);

*(even stronger) the diagonal functor $C_0\to C^{\{\bullet\bullet\}}$ has a right adjoint?

(Here $\{\bullet\bullet\}$ denotes the discrete category with two objects and only identity arrows.)
 A: One good strategy for searching for counterexamples in category theory is to look at the example of preorders (i.e., categories with at most one arrow between any two objects). Or, even simpler, posets.
To come up with an example satisfying your criteria, we want to find an example of a poset $C$ with a sub-poset $C_0$ such that:

*

*There are $a,b\in C_0$ with no greatest lower bound in $C$.

*For all $x,y\in C$, there exists $z\in C_0$ such that for all $w\in C_0$, $w\leq z$ if and only if $w\leq x$ and $w\leq y$. That is, $x$ and $y$ have a lower bound in $C_0$ which is greatest among lower bounds in $C_0$.

Let $C = \mathbb{N}\cup \{\infty_1,\infty_2\}$. The order relation is the usual one between elements of $\mathbb{N}$, we have $n\leq \infty_1$ and $n\leq \infty_2$ for all $n\in \mathbb{N}$, and $\infty_1$ and $\infty_2$ are incomparable.
Let $C_0 = \{0,\infty_1,\infty_2\}\subseteq C$.

*

*$\infty_1$ and $\infty_2$ are in $C_0$ and have no greatest lower bound in $C$ (the set of common lower bounds is $\mathbb{N}$, which has no greatest element)

*Let $x,y\in C$. If $x = y = \infty_i$ for $i \in \{1,2\}$, let $z = \infty_i\in C_0$. This is the greatest lower bound in $C_0$ (even in $C$!) for $x$ and $y$. Otherwise, the only lower bound for $x$ and $y$ in $C_0$ is $z = 0$, so this is the greatest lower bound for $x$ and $y$ in $C_0$.

