necessarily covariant vs necessarily contravariant vs neither

Question

From Vakil's online notes on algebraic geometry he says there exists a contravariant functor from TOP to RING taking a topological space X to the ring of continuous functions on X, whereby a continuous map from X to Y induces a pullback from continuous functions on Y to continuous functions on X. My question is: This functor needn't be contravariant right? We could have just as legitimately defined the same thing except covariant? Does the same apply when we consider the category of vector spaces with the functor of taking duals? Does that NEED to be contravariant? Further illuminating examples of the difference between functors which CAN be both covariant and contravariant and those which can only be one are welcome. Thanks

Answer 1

As long as the ring $C(X)$ that you associate to the topological space $X$ is the ring of continuous functions, this association (i.e. functor) is necessarily contravariant. This is because if $F : X \to Y$ is a continuous map (i.e. a morphism in $TOP$), then the natural morphism to consider between the corresponding objects in $RING$ is $C(F) : C(Y) \to C(X)$ defined by $C(F) (f) = f \circ F$. Given that this functor reverses the arrows (i.e. $F$ goes from $X$ to $Y$, while $C(F)$ goes in the opposite way), this is necessarily contravariant, so you don't have any freedom of choice.

Of course, if instead of attaching $C(X)$ to $X$ you attached something else, for instance $C(X)^*$ (the algebraic or topological dual), then this new functor would become covariant. You have to specify first your functor, and its covariance or contravariance will follow without you having to impose it.