Functors

Maps between categories: objects and arrows carried across, composites intact, sometimes with the arrows turned around.

The idea

A functor is a map between categories. It sends objects to objects and arrows to arrows, and it respects the only structure a category carries: composition and identities.

Definition (Functor).

A functor $F \colon \mathcal{C} \to \mathcal{D}$ assigns to each object $A$ of $\mathcal{C}$ an object $F(A)$ of $\mathcal{D}$, and to each arrow $f \colon A \to A'$ an arrow $F(f) \colon F(A) \to F(A')$, subject to two axioms: $F(g \circ f) = F(g) \circ F(f), \qquad F(1_{A}) = 1_{F(A)},$ the first for every composable pair $f \colon A \to A'$, $g \colon A' \to A''$ and the second for every object $A$.

Composites go to composites and identities go to identities.

One example shows the axioms at work. Send a set $X$ to its set of subsets $P(X)$, and a function $k \colon X \to Y$ to the direct-image map carrying a subset $S$ to $\{k(s) : s \in S\}$. This is a functor from $\mathbf{Set}$ to itself: the image of a subset under a composite is the image of its image, and the image under an identity is the subset itself.

Some constructions reverse direction. Send the same $k$ to the preimage map instead and it runs from $P(Y)$ to $P(X)$, so a composite can only be undone outermost first and the factors compose in the opposite order. Such a construction is contravariant, always for the same reason: its value is built out of maps out of the object, so the only way to use an arrow is to precompose with it.

Ways to work on it

Not sure where to start? Take the ten-question placement test.