Why is $\mathbf{Rel} \cong \mathbf{Rel}^{\mathbb{op}}$? I haven't yet fully grasped Category Theory so I am doing the exercises in Awodey's book for the first chapter, and exercise 2a) is confusing me very much.
The question is to prove or disprove that $\mathbf{Rel} \cong \mathbf{Rel}^{\mathbf{op}}$. I get that the idea is 
$$f^{\mathbf{op}}: D \longrightarrow C = \{\langle d,c \rangle \in D\times C| (c,d) \in f\}$$
But I can't write the functor because according to the definition: $$F(f:C\longrightarrow D)=F(f):F(C)\longrightarrow F(D)$$
So if there are two relations from $C$ to different objects, $F(C)$ would not be defined.
What am I doing/understanding wrong? Thank you!
edit: I should add that I say $\mathbf{Rel} \cong \mathbf{Rel}^{\mathbb{op}}$ not because I proved it but because I read a document with the solutions...
 A: One of the problems with these exercises in Awodey is that he didn't actually define "isomorphism of categories" in the first chapter. But let us take it to mean "isomorphism in $\sf Cat$", the category of (small) categories and functors.
In the preceding exercise, one is asked to show that:
$$C: {\sf Rel}^{\rm op} \to {\sf Rel}$$
given as the identity on objects, and on relations as $C(R^{\rm op}: D^{\rm op} \to C^{\rm op}) = R^c: D \to C$ (note that $R^{\rm op}:D^{\rm op} \to C^{\rm op}$ corresponds to $R: C \to D$ in $\sf Rel$) is a functor.
Let us check the composition property of $C$ on $S^{\rm op}: E^{\rm op}\to D^{\rm op}$ and $R^{\rm op}$:
\begin{align}
C(S^{\rm op}R^{\rm op}) &= C((RS)^{\rm op}) & &\text{Definition of composition in $\sf Rel^{\rm op}$}\\
&= (RS)^c & &\text{Definition of $C$}\\
&= \{(e,c) \mid (c,e) \in RS\} &&\text{Definition of $(-)^c$}\\
&= \{(e,c)\mid \exists d \in D: (c,d)\in R, (d,e)\in S\} &&\text{Definition of $RS$}\\
&= \{(e,c)\mid \exists d \in D: (d,c)\in R^c, (e,d) \in S^c\}\\
&= S^c R^c = C(S^{\rm op})C(R^{\rm op})
\end{align}
It's not hard now to verify that $C$ is indeed a functor. I'm sure you can define a functor $C^{\rm op}: {\sf Rel}\to {\sf Rel}^{\rm op}$ that together with $C$ gives ${\sf Rel} \cong {\sf Rel}^{\rm op}$.
As a general advice, I suggest that you meticulously write the $\rm op$ superscript on every object and morphism in an opposite category until you know your way around. This can help to overcome some of the initial confusion.
