Definition of a Contravariant functor Def: A contravariant functor between categories $C$ and $D$ contains the same data as a functor $F:C\rightarrow D$, except 
$$
     F(g \circ f) = F(f) \circ F(g)
$$
Under this "definition" (I'm reading a text from a physics perspective), it seems like a contravariant functor is not a functor, despite what the name suggests, since functors between categories have to commute with composing morphisms, whereas these anticommute. Is this an accurate statement? 
Also, it seems that this definition requires the categories in question to have at least 3 objects, in order to even test if a functor between them is contravariant. So for example, something that I would expect to be a contravariant functor, 
$$
A \rightarrow B\ \ \  \mapsto \ \ \ \ A \leftarrow B
$$
it's not clear to me if we can say that functor is contravariant under the given definition. 
 A: If you define a separate notion of "contravariant functor", then a contravariant functor is not a functor. You have two separate notions then: covariant functors from $\mathcal C$ to $\mathcal D$ and contravariant functors from $\mathcal C$ to $\mathcal D$.
Personally, I think having a notion of "contravariant functors" is a very bad idea. Instead, as Lord Shark the Unknown states, we can view the phrase "a contravariant functor from $\mathcal C$" as meaning "a (covariant) functor from $\mathcal C^{op}$". Then there is just a single notion of functor corresponding to the covariant case. With this approach, it never makes sense to say something like "a contravariant functor $F:\mathcal C^{op}\to\mathcal D$". You would say "a functor $F:\mathcal C^{op}\to\mathcal D$". This approach makes the $\mathcal C\to\mathcal D$ unambiguous and self-contained. With the co-/contra-variant functor approach, the $\mathcal C\to\mathcal D$ notation becomes ambiguous. I also find it useful to have the ${}^{op}$s explicit as you can make sure that compositions "type-check". On the other hand, this explicit information is usually inferrable and thus is arguably adding noise.
A: I think it’s helpful to write out what the contravariant functors do to the objects.
Let $F:C\rightarrow D$ be a contravariant functor, and $f:X\rightarrow Y, g:Y\rightarrow Z$ be morphisms in $C$.
Note that $F$ reverses morphisms: $F(f:X\rightarrow Y) = Ff:FY\rightarrow FX$ and $F(g:Y\rightarrow Z) = Fg:FZ\rightarrow FY$.
This makes the equation work.
