What constructions of "elementary" mathematics are actually functors?

Question

I'm not looking for the usual simple examples of functors like the fundamental group or forgetful functors, what I'm looking for is some interesting examples of constructions from "elementary" mathematics that are secretly functorial. Like the derivative: it's actually a functor, with the chain rule expressing the composition rule of functors, but that's never discussed in basic Calculus courses.

Answer 1

One of my favourite examples of these is group actions.

A (monoid or) group $G$ can be considered as a category with a single object $\star$, whose morphisms $\star \to \star$ are the elements of $G$, and whose identity and composition are given by the unit element $e$ and the group operation, respectively.

A left action of $G$ on a set $X$ is precisely a functor $\alpha : G \to \mathbf{Set}$, where $G$ is considered as a category in the above sense.

• The set $X$ is the value of $\alpha(\star)$;
• For each $g \in G$, we obtain a function $\alpha_g = \alpha(g) : X \to X$;
• Functoriality expresses the fact that $\alpha_e = \mathrm{id}_X$ and $\alpha_{gh} = \alpha_g \circ \alpha_h$.

Likewise a right action on a set is precisely a functor $\alpha : G^{\mathrm{op}} \to \mathbf{Set}$.

Answer 2

In almost any course a (U.S.) undergraduate math major would take, the construction $F(X)=A \times X$ for fixed $A$ defines a functor from any reasonable category to itself (obvious action on maps). This works for sets, (abelian) groups, topological spaces, vector spaces, rings, etc. (Of course it works in any category with products, but I'm trying to keep it "elementary.")

Answer 3

As noted, differentiation is a functor in the category of (real, finite-dimensional) smooth manifolds. In particular, it maps every manifold $X$ to its tangent bundle $TX$, which is locally isomorphic to $X \times \mathbb{R}^{n}$ for appropriate $n$. In some sense, one might consider differentiation ''locally'' an example functor of the form $F(X) = A \times X$ described by @Randall (well, technically $X \mapsto X \times A$).

However, the functor maps smooth functions in a non-trivial way. For manifolds $X, Y$ and smooth $f: X \to Y$, the derivative $D(f): TX \to TY$ is the map $$ D(f): (x,v) \mapsto (f(x), df_{x}(v)) $$ where $df_{x}$ is the ordinary total derivative of $f$ at $x$.

Answer 4

The function that sends a set $X$ to its powerset $\mathcal{P}(X)$ is a functor; and interestingly, it is a functor in more than one way!

Probably the most natural way to make it into a functor is to define, given a function $f:X\to Y$, $$\mathcal{P}(f):\mathcal{P}(X)\to \mathcal{P}(Y):A\mapsto f(A)=\{f(a)\mid a\in A\};$$ in other words, we send $f$ to the function "direct image by $f$". This is a functor, because (pretty much by definition) $g(f(A))=(g\circ f)(A)$ for all $f:X\to Y$, $g:Y\to Z$ and $A\subset X$.

But there is also the function "inverse image by $f$", defined as $$\mathcal{P}'(f):\mathcal{P}(Y)\to \mathcal{P}(X):B\mapsto f^{-1}(B)=\{a\in X\mid f(a)\in B\}.$$ Note that here I've switched $X$ and $Y$; so it's not a functor on the category of sets, but from the opposite category of sets to the category of sets, or if you prefer a contravariant functor on the category of sets. Here the functoriality amounts to the fact that $f^{-1}(g^{-1}(C))=(g\circ f)^{-1}(C)$ for all $f:X\to Y$, $g:Y\to Z$ and $C\subset Z$.

Answer 5

Any homomorphism $f:G_1\rightarrow G_2$ between two groups, both considered as one object categories, is a functor from $G_1$ to $G_2$.

This is true for ring homomorphisms as well.