# What is a Natural Transformation? Definition and Examples, Part 2

/Continuing our list of examples of natural transformations, here is...

## Example #2: double dual space

*until*we pick a basis.*

On the other hand, there is an isomorphism $V\overset{\cong}{\longrightarrow}V^{**}$ that requires

*no*choice of basis: for each $v\in V$, let $\text{eval}_v:V^*\to\mathbb{k}$ be the

*evaluation map*. That is, whenever $f:V\to \mathbb{k}$ is an element in $V^*$, define $\text{eval}_v(f):=f(v)$. Folks often refer to this isomorphism as

**natural**. It's natural in the sense that it's there for the taking---it's patiently waiting to be acknowledged, irrespective of how we choose to "view" $V$ (i.e. irrespective of our choice of basis). This is evidenced in the fact that $\text{eval}$ does the

*same*job on

*each*vector space throughout

*entire*category. One map to rule them all.**

For this reason, the totality of all the evaluation maps assembles into a natural transformation (a natural

*isomorphism*, in fact) between two functors!

To see this, let $(-)^{**}:\mathsf{Vect}_{\mathbb{k}}\to\mathsf{Vect}_{\mathbb{k}}$ be the the

*double dual*functor $(-)^{**}$ that sends a vector space $V$ to $V^{**}$ and that sends a linear map $V\overset{\phi}{\longrightarrow}W$ to $V^{**}\overset{\phi^{**}}{\longrightarrow} W^{**}$, where $\phi^{**}$ is precomposition with $\phi^{*}$ (which we've defined before). And let $\text{id}:\mathsf{Vect}_{\mathbb{k}}\to\mathsf{Vect}_{\mathbb{k}} $ be the identity functor.

Now let's check that $\text{eval}:\text{id}\Longrightarrow (-)^{**}$ is indeed a natural transformation.

As per our discussion last time, this suggests that $\text{id}$ and $(-)^{**}$ are really the same functor up to a change in perspective. Indeed, this interpretation pairs nicely with the observation that any vector $v\in V$ can either be viewed as, well, *a vector*, or it can be viewed as an assignment that sends a linear function $f$ to the value $f(v)$. In short, $V$ is genuinely and authentically just like its double dual.

They are - quite naturally - isomorphic.

## Example #3: representability and Yoneda

**representable**if, loosely speaking, there is an object $c\in\mathsf{C}$ so that for all objects $x$ in $\mathsf{C}$, the elements of $F(x)$ are "really" just maps $c\to x$ (or maps $x\to c$, if $F$ is contravariant). As an illustration, we noted that the functor $\mathscr{O}:\mathsf{Top}^{op}\to\mathsf{Set}$ that sends a topological space $X$ to its set $\mathscr{O}(X)$ of open subsets is

*represented by*the Sierpinski space $S$ since $$\mathscr{O}(X)\cong \text{hom}_{\mathsf{Top}}(X,S)$$ where I'm using $\cong$ to denote a set bijection/isomorphism. So in other words, an open subset of $X$ is essentially the same thing as a continuous function $X\to S.$ (We discussed this in length here.)

*natural!*That is, the ensemble of isomorphisms $\mathscr{O}(X)\overset{\cong}{\longrightarrow}\text{hom}_{\mathsf{Top}}(X,S)$ (one for each $X$) assemble to form a natural isomorphism between the two functors $\mathscr{O}$ and $\text{hom}_{\mathsf{Top}}(-,S)$.***

**representable**if there is an object $c\in\mathsf{C}$ so that $F$ is

*naturally isomorphic*to the hom functor $\text{hom}_{\mathsf{C}}(c,-)$, i.e. if $$F(x)\cong\text{hom}_{\mathsf{C}}(c,x) \qquad \text{naturally, for all $x\in \mathsf{C}$}$$ (or if $F$ is contravariant, $F(x)\cong\text{hom}_{\mathsf{C}}(x,c)$).

Here's a very simple example. Suppose $A$ is any set and let $*$ denote the set with one element. Notice that a function from $*$ to $A$ has exactly one element in its image, i.e. the range of $*\to A$ is $\{a\}$ for some $a\in A$. This suggests that a map $*\to A$ is *really* just a choice of element in $A$! Intuitively then, the elements of $A$ are in bijection with functions $*\to A$,
$$A\cong\text{hom}_{\mathsf{Set}}(*,A).$$
But more is true! The isomorphism $A\to \text{hom}_{\mathsf{Set}}(*,A)$ which sends $a\in A$ to the function, say, $\bar{a}:*\to A$, where $\bar{a}(*)=a$, is natural. That is, for any $A\overset{f}{\longrightarrow}B$, the following square commutes

*then*think of $f(a)$ as a map $*\to B$. OR first think of $a$ as a map $*\to A$, and then postcompose it with $f$.

In short, the identity functor $\text{id}:\mathsf{Set}\to\mathsf{Set}$ is represented by the one-point set $*$ since every function $*\to A$ is really just a choice of an element $a\in A$.

To whet your appetite, I'll quickly say that one consequence of the Lemma is that we are prompted to think of an object $x$ -- no longer as an object, but now -- as a (representable) functor $\text{hom}(x,-)$, similar to how we may think of a point $a\in A$ as a map $*\to A$.

And *that* is The Most Obvious Secret of Mathematics!

*One can show that there is *no *"natural" isomorphism of a vector space with its dual. For instance, see p. 234 of Eilenberg and Mac Lane's 1945 paper, "The General Theory of Natural Equivalences."

*all*$x\in X$.

***Here, $\text{hom}_{\mathsf{Top}}(-,S):\mathsf{Top}^{op}\to\mathsf{Set}$ is the contravariant functor that sends a topological space $X$ to the set $\text{hom}_{\mathsf{Top}}(X,S)$ of continuous functions $X\to S$ and that sends a continuous function $X\overset{f}{\longrightarrow} Y$ to its pullback $\text{hom}_{\mathsf{Top}}(Y,S)\overset{f^*}{\longrightarrow}\text{hom}_{\mathsf{Top}}(X,S).$