7
$\begingroup$

There is a bijection between $\mathcal P(X)$ and $2^X$ where $2=\{0,1\}$, namely $U\subseteq X\mapsto f$ where $f(x)=1$ iff $x\in U$. This bijection is “natural” in an informal sense.

On the other hand, there may be a bijection between $\mathcal P(X)$ and some other arbitrary set $Z$, that happens to have the same cardinality. I would call this bijection informally “unnatural”.

Does my informal idea of “natural” in this context correspond to an idea of naturality in category theory?

$\endgroup$
5
$\begingroup$

In this context, yes, because the bijection between $\mathcal{P}(X)$ and $2^X$ is natural category-theoretically (it's usually an exercise you do when you are learning category theory), while if you just take $Z$ of the same cardinality, then naturality doesn't even make sense because on one side you have a functor ($X\mapsto \mathcal{P}(X)$) and on the other side you have a single object $Z$, so it doesn't even make sense to speak of naturality.

However if you restrict to a single $X$, then there are two categories you may want to have a look at : the trivial category on $X$, that has only $id_X$ as morphism, or the full subcategory of $\mathbf{Set}$ on $X$, which has as morphisms $\hom(X,X)$. In the first case, the functors $X\mapsto \mathcal{P}(X)$ and the constant functor $X\mapsto Z$ will be naturally isomorphic; in the second case, the functor $X\mapsto \mathcal{P}(X)$ acting the usual way on maps, and the constant functor $X\mapsto Z$ aren't naturally isomorphic - though there is a way to act on arrows to have a functor $X\mapsto Z$ that is naturally isomorphic to $X\mapsto \mathcal{P}(X)$.

What this shows is that to make sense of what "natural" means you have to be careful about what the categories involved are; and how the functors act, not only on objects, but also on arrows.

$\endgroup$
4
$\begingroup$

Here's a possible interpretation: take the contravariant functor

$$ \newcommand{\c}[1]{\mathbf{#1}} \newcommand{\cop}[1]{\c{#1}^{op}} \newcommand{\parts}[1]{\mathcal{P}(#1)} \newcommand{\2}[1]{2^{#1}} \begin{align} \parts{-} : &\ \c{Set} \longrightarrow \cop{Set} \\ & X \longmapsto \parts{X} \\ & \downarrow^f \ \mapsto \ \uparrow^{\parts{f}} \\ & Y \longmapsto \parts{Y} \end{align} $$

with $\parts{f}(A) = f^{-1}(A)$. Using your notation, there is another contravariant functor in $\c{Set}$ that can be defined as $\2{-} := \hom_{\c{Set}}(-,\{0,1\})$. To formalize your intuition, one would like to have a natural transformation between these.

In fact, we actually have a natural isomorphism defined by: $$ (\eta_X : \parts{X} \to 2^X)_{X \in \c{Set}}, \quad \eta_X(A)(x) = \cases{1 \text{ if } x \in A \\ 0 \text{ otherwise}} $$

which is to say that $\eta_X(A) = \chi_A \in \hom(X,\{0,1\}) = \2{X}$. In effect, $\eta$ is natural because given an arrow $f :X \to Y$ in $\c{Set}$, $$ \eta_X\parts{f}(A) = \eta_X(f^{-1}(A)) = \chi_{f^{-1}(A)} = \chi_Af = f^*(\chi_A) = \2{f}(\eta_X(A)) $$

with $2^f = f^*$ being the precomposition of $f$, i.e. $f^*(g) := gf$. Finally, $\eta$ is an isomorphism as each arrow $\eta_X$ is

  • injective because if $A \neq A' \in \parts{X}$ then $\eta_X(A)(z) \neq \eta_X(A')(z)$ for $z \in A' \triangle A$.
  • surjective because given $s : X \to 2$, then $s = \chi_{s^{-1}(1)} = \eta_X(s^{-1}(\{1\}))$.

and isomorphisms in $\c{Set}^{op}$ are bijections.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.