Natural Transformation¶

Natural transformations are, along with the category and the functor, a fundamental concept in Category Theory. A natural transformation is a mapping from one functor to another with the same domain and codomain as the first, subject to a naturality constraint.

In particular, let $C$ and $D$ be categories, and let $F, G: C \to D$. A natural transformation $\alpha: F \Rightarrow G$ is a map from $F$ to $G$ such that for each object $c \in C$, there is an arrow, called a component of the natural transformation, $\alpha_c: Fc \to Gc$, such that the [[Square Diagram]] commutes[^1].

Vector Spaces and Linear Duals¶

The most common motivating example for the natural transformation is that of vector spaces and their linear duals. Let $V$ be a finite-dimensional vector space over the field $k$. It admits a linear dual space $V^* = \text{Hom}(V, k)$ of linear maps from $V$ to $k$. (This is to say, it is the Hom-Set between $V$ and $k$ in the category of vector spaces.)

We can show that $V$ is isomorphic to $V^*$ by constructing a dual basis. Let $e_i \in V$ define a basis on $V$, and let $e_i^* \in V^*$ such that $e_i^* e_j = \delta_{ij}$. Then $e_i^*$ define a basis on $V^*$, and the map between the basis defines an isomorphism. However, it is somewhat unsatisfying that we must adopt a particular basis in order to arrive at this conclusion.

In a more categorical view, we have a dual-space functor $(-)^*: \text{Vect}_k^{\text{op}} \to \text{Vect}_k$. This is a contravariant functor because linear maps will be reversed when switching to the dual description. For instance, mappings $\mathbb{R}^2 \to \mathbb{R}^3$ are $3 \times 2$ matrices, but these are mapped to $2 \times 3$ matrices under the dual-space functor.

Pushing this further, we can consider the double-dual $V^{**} = \text{Hom}(\text{Hom}(V, k), k)$ and the double-dual functor $(-)^{**}: \text{Vect}_k \to \text{Vect}_k$. We can certainly construct this by composing the dual-space functor with itself, but there is a more "natural" construction.

Define the evaluation function $f(v): V^* \to k$, which maps an element $u^* \in V^*$ to $u^* v \in k$. Given an element $v \in V$, this is a mapping from $V^*$ to $k$, which means that the evaluation function is an element of $\text{Hom}(V^*, k) = \text{Hom}(\text{Hom}(V, k), k) = V^{**}$ for each $v \in V$. That is to say, the evaluation function generically is a mapping $V \to V^{**}$. This mapping is in fact an isomorphism, and so $V \cong V^{**}$ without the need to refer to a basis.

The fact that the process of showing that $V \cong V^*$ required an appeal to a particular basis but showing that $V \cong V^{**}$ did not reflects that the latter is a natural isomorphism, whereas the former is not.

Examples of Natural Transformations¶

There are many further examples of natural transformations, but some especially interesting ones are listed here.

There exists a pair of functors from $\text{Top}$ to $\text{Set}$ which map topological spaces to the set of their open sets or closed sets. While topologists generally opt to speak about open Sets as the natural framework for topology, it is well-known that all theorems could equally be converted into statements about closed sets. This fact is a consequence of the fact that the two functors are not only isomorphic, but naturally isomorphic.
Consider the category whose objects are [[Hilbert Space]]s and whose morphisms are linear operators between them, and an endofunctor on this category such that a Hilbert space $\mathcal{H}$ is mapped to $\mathcal{H} \otimes \mathcal{H}$. It turns out that there is only one natural transformation from the identity endofunctor to this functor: the one whose components are the zero maps. This is a categorical statement of the [[No-Cloning Theorem]].
Let $A$ and $B$ be sets, let $\times$ and $+$ denote their Cartesian product and disjoint union (respectively), and let $A^B$ denote the set of functions from $B$ to $A$. Then the following are natural isomorphisms:

$$ A \times (B + C) \cong (A \times B) + (A \times C), \quad (A \times B)^C \cong A^C \times B^C $$ $$ A^{B+C} \cong A^B \times A^C, \quad \left(A^B\right)^C \cong A^{B \times C} $$ reflecting the standard arithmetic relations. There exists a functor from the category of sets to the category of partially-ordered real numbers which makes this analogy manifest.

[^1] This wiki does not yet support graphical commutative diagrams, but they will hopefully be introduced in an update soon.