Cartesian Closedness of symmetric binary relations as reflective subcategory of Rel

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.

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.
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.
• I want to prove that ${\bf Sym}$ is an exponential ideal (and hence to conclude by Johnstone+Adamek that it is cartesian closed). But as S.C. answered, there is a mistake in Adamek. So, in order to show that ${\bf Sym}$ is cartesian closed, I must prove it directly (using the fact that inherits the Cartesian closedness as you showed). Aug 21, 2021 at 19:23