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Looking at the category $\mathbf{Rel}$ and its opposite, I would like to know if there is something I'd call identity functor, $f:\mathbf{Rel} \to \mathbf{Rel}^{OP}$ that sends a set to itself and also every relation to itself. If you have a relation $R\subseteq A\times B$, you could say that it is also a morphism of $\mathbf{Rel}^{OP}$. So to my mind, $f$ is an isomorphism between $\mathbf{Rel}$ and $\mathbf{Rel}^{OP}$ because it respects concatenation of arrows.

The definition of arrows $f:A\to B$ in the opposite category just says that they are arrows $f:B\to A$ in the category itself. Now I do not know whether I can say: "The functor is well-defined because there is a corresponding relation $R'\subseteq B\times A$".

I doubt but hope that my question is understandable.

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    $\begingroup$ Yes, $\mathbf{Rel}$ is isomorphic to its opposite via the functor you describe. It may be a strange fact at first, but it is true. $\endgroup$ – Zhen Lin Feb 20 '15 at 16:16
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This is not called a identity functor. It maps a relation $R \subseteq A \times B$ to the relation $R^{\dagger} \subseteq B \times A$ defined by $R^{\dagger} := \{(b,a) : (a,b) \in R\}$. In fact, this defines a dagger-structure on $\mathsf{Rel}$. See Wikipedia for some information about dagger categories.

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