# How exactly does a map represent an operation?

My question is related to exercise 5.7 in Sets for Mathematics by Lawvere and Rosebrugh (p. 105). We are given a map $\beta: B_2 \to B_1$ and a contravariant functor $2^{\beta}: 2^{B_1} \to 2^{B_2}$ constructed via exponentiation in the following manner:

Having $\beta$ and the characteristic map $\phi: B_1 \to 2$ we get an exponential object $2^{B_1}$ -- this follows from the Axiom of Exponentiation. We also get an evaluation mapping $eval: 2^{B_1} \times B_1 \to 2$. Since there is a composed map $\phi \circ \beta : B_2 \to 2$, there is another exponential object $2^{B_2}$. Here follows the trick: we construct a product mapping $1 \times \beta : 2^{B_1} \times B_2 \to 2^{B_1} \times B_1$; compose it with $eval$ and get $eval \circ 1 \times \beta : 2^{B_1} \times B_2 \to 2$; by the Axiom, this is all we need to construct a function space, hence $2^{\beta} : 2^{B_1} \to 2^{B_2}$.

Now. We are asked to prove that $2^{\beta}$ represents "the operation of taking the inverse image along $\beta$ of arbitrary parts of $B_1$". All of the concepts used are pretty clear to me. Except for one.

It is not clear to me what exactly the word "represents" means in this case. That is why I am not asking you to solve the problem for me. I'd be extra grateful if someone provided an example of some other map representing an operation.

P. S. First post here. Sorry, little experience with formulating maths questions. Bear with me, I'll improve.

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This might be obvious to category theorists, but what do you mean by "constructed via exponentiation"? – Karl Kronenfeld Apr 8 '13 at 10:24
You can read "represents" naïvely here in the English sense, but it also "represents" in the technical sense of representable functors. (Recall that the object $2$ is a subobject classifier for $\textbf{Set}$.) – Zhen Lin Apr 8 '13 at 10:36
@user1, I added an explanation. My background is not in maths so my presentation skills may be impoverished. Still I hope this helps. – Haec Apr 8 '13 at 14:17

As Zhen Lin commented, primarily the word "represents" in this context could be just replaced by "is"

$2^\beta$ is the operation of taking inverse image along $\beta$.

On the other hand, this contravariant powerset functor is naturally isomorphic to the hom-functor $A \mapsto \hom(A,2)$ (where $2=\{0,1\}$), and that means that

The object $2$ represents the contravariant powerset functor.

In addition, many kinds of operations can be represented by morphisms, e.g. a binary operation on a set $A$ is nothing but a morphism $A\times A\to A$ in the category $\mathcal{Set}$. This is associative iff the two arising morphisms $A\times A\times A\to A$ are identical. Now, for example, if we consider instead the category $\mathcal{Ab}$ of Abelian groups and $\otimes$ instead of $\times$, an additional associative operation (in the above sense) on a given object (Abelian group) makes it a ring..

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It should be "contravariant", not "covariant". – Zhen Lin Apr 8 '13 at 12:26
Thank you @Berci! – Haec Apr 8 '13 at 13:43
@ZhenLin: Yes, of course. Corrected. – Berci Apr 8 '13 at 14:48