Cartesian Closedness of symmetric binary relations as reflective subcategory of Rel

Question

In Adamek's book "The Joy of Cats", the category ${\bf Rel}$ of pairs $(\Omega,\rho)$ (with $\rho$ any binary relation on $\Omega$) and relation-preserving maps has been studied. The authors prove that ${\bf Rel}$ is cartesian closed by computing the exponential objects, namely if $(A,\rho),(B,\sigma) \in {\bf Rel}$, their power object is given by $(B^A,\tau)$, where $f \tau g \iff [(x,y) \in \rho \implies (f(x),g(y)) \in \sigma]$.

Next, they say that the full subcategory ${\bf Sym}$ of symmetric binary relations is reflective (and it is ok!) and also cartesian closed because the reflector preserves finite products. However, in Proposition $A4.3.1$ of Johnstone's Sketch of an Elephant, when $\mathfrak{C}$ is a cartesian closed category and $\mathfrak{D}$ a full reflective subcategory, the reflector preserves finite products if and only if $\mathfrak{D}$ is an exponential ideal (i.e. $Y \in \mathfrak{D},X \in \mathfrak{C} \implies Y^X \in \mathfrak{D})$. But, now, I'm trying to verify that ${\bf Sym}$ is an exponential ideal. I computed the power object of $(A,\rho) \in {\bf Rel}$ and $(B,\sigma) \in {\bf Sym}$ in ${\bf Rel}$. The corresponding power object $(B^A,\tau)$ seems not to belong to ${\bf Sym}$. In fact, if $f,g: A \longrightarrow B$ are any two functions such that $(f,g) \in \tau$ (i.e. $(x,y) \in \rho \implies (f(x),g(y)) \in \sigma$), in general it is not true that $(g,f) \in \tau$. Who ensures me that if $(x,y) \in \rho$, then $(g(x),f(y)) \in \sigma$? In general $(y,x) \notin \rho$ so I cannot use the symmetry of $\sigma$. So, in order to solve the issue, I must take $Hom_{\bf Rel}((A,\rho),(B,\sigma))$ instead of $B^A$. Where is the error in my argument?

P.S. The fact that ${\bf Sym}$ is cartesian closed may be easily seen taking $Hom_{\bf Rel}((A,\rho),(B,\sigma))$ instead of $B^A$. In this case, all is right! And my question arises because of the above post scriptum. If Adamek is not in error, then I expect that $B^A=Hom_{\bf Rel}((A,\rho),(B,\sigma))$, which is clearly false.

Answer 1

The reflector (which is written out in 4.17.A, if you want to check) indeed does not preserve products, so that seems to be a mistake in Joy of Cats. The product is effectively the conjunction of the two relations and the reflector is effective the disjunction of a relation and its inverse. Thus, for the reflector to preserve products, we'd need a logical equivalence $(P$ and $Q)$ or $(R$ and $S) \iff (P$ or $R)$ and $(Q$ or $S)$, but the implication from right to left doesn't hold.

For a definite counterexample, consider the set with two objects $x$ and $y$ where $x$ is related to $y$ but not vice versa. See if the reflector preserves the product of this relation with itself.

What may have been meant is that the inclusion functor preserves products and the exponential of two symmetric binary relations is again symmetric. Those two facts, together with the facts that the subcategory has finite products and the inclusion functor is full imply that the subcategory is cartesian closed if the full category is. That the inclusion functor preserves products follows from the existence of the reflector (which also shows that the inclusion preserves all limits), but is easy enough to show directly starting from the definitions of the products in each category.

Answer 2

The exponential object in $\mathbf{Sym}$ is the same as the one in $\mathbf{Rel}$. This is because if $(f,g) \in \tau$ and $(x,y) \in \rho$, then $(y,x) \in \rho$ (because $\rho$ is symmetric), hence $(f(y),g(x)) \in \sigma$ (by the definition of $\tau$), hence $(g(x),f(y)) \in \sigma$ (because $\sigma$ is symmetric). This means that $(g,f) \in \tau$, so $\tau$ is symmetric.