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A functor assigns an object of a category $F(A)$ to each object $A$ and a categorical morphism $F(f):F(A)\to F(B)$ to each arrow $f:A\to B$. It preserves the rules for composing arrows:For example, the forgetful functor from the category of groups to the Category of sets keeps the underlying sets and maps of elements while forgetting the group operations. A group homomorphism remains a function, and composing the underlying functions gives the underlying map of the composite group homomorphism.

Another example sends a set $X$ to its power set $\mathcal{P}(X)$. It sends a function $f:X\to Y$ to the function $\mathcal{P}(f):\mathcal{P}(X)\to \mathcal{P}(Y)$ defined by $S\mapsto f(S)$, the image of the subset $S$. For $f(x)=x+1$ and $g(x)=2x$, the subset $\{0,2\}$ goes to $\{1,3\}$ and then $\{2,6\}$; applying the composite $g\circ f$ directly also gives $\{2,6\}$. The equality of these routes expresses preservation of composition in a category.

A natural transformation between functors $F,G:\mathcal{C}\to \mathcal{D}$ assigns a categorical morphism ${\alpha }_A:F(A)\to G(A)$ to every object of a category $A$, withThis naturality condition says that transforming the input before transporting it gives the same result as transporting it first and transforming the output. It constrains the whole family of arrows, so unrelated choices of a map for each object generally fail it.

For a concrete example, compare the functor that leaves every set and function unchanged with the power set functor using images of subsets. The maps ${\eta }_X:X\to \mathcal{P}(X)$ given by $x\mapsto \{x\}$ form a natural transformation: for any function $f:X\to Y$,The right-hand panel of the category illustration checks the square at $x=3$ with $f(x)=x+1$: both routes end at $\{4\}$. The statement works for every function, rather than depending on a preferred labeling of a particular set.

Let $C\mathcal{A}={\pi }_0\mathcal{A}$ be the set of connected components of a category: two objects of a category are identified when joined by a finite zigzag of morphisms. Let $DA$ be the discrete category on $A$, and let $IA$ be the indiscrete category on $A$, with exactly one morphism for every ordered pair of objects.

A functor $\mathcal{A}\to DA$ must give the same object value along every arrow, hence along every zigzag. Conversely a function ${\pi }_0\mathcal{A}\to A$ uniquely defines such a functor. A functor $DA\to \mathcal{A}$ is just a choice of an object of a category for each element of $A$. A functor $\mathcal{A}\to IA$ is likewise determined by an arbitrary object function, since its arrow images are forced and always compose correctly. ThereforeThis is the adjoint chain for the objects of a category: