Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When dealing with a space $X$, that posses a lot of structure (complete lattice, complete metric space, vector space), what can be said about the cartesian exponentiation $X^Y=\{f \mid f:Y\rightarrow X\}$ (whic is the set of functions from another set $Y$ to $X$)?

It is intuitivly clear, that the point-wise versions of operations and relations in $X$ carry over many properties of $X$ to $X^Y$. (e.g. $X^Y$ is also a vector space with point-wise addition and multiplication, it is also a complete lattice with the point-wise or product order. It is also a complete metric space under the maximum (or point-wise) metric).

Is there a field of mathematics, that is concerned with this particular sharing of structure between a space $X$ and its function space $X^Y$?

Or more pragmatically, what results can I use, in order to spare me the proof of these 16 mostly trivial and shallow implications (8 for vector space, 3 for partial order + 1 completeness, 3 for metric + 1 completemess)?

share|cite|improve this question
Do you mean the set of all set-theoretic functions $Y \to X$, or the relevant notion of structure-preserving maps? One of these questions is significantly easier than the other. – Zhen Lin Mar 12 '13 at 16:53
@ZhenLin I am talking about the set of all functions from $Y$ to $X$. (I hope this is the easier one!). I could not wrap my head around, how this might be related to structure-preserving maps. – JSG Mar 12 '13 at 17:15
So, this is a "product" ... it is indexed by $Y$ and all factors are the same. But in all of your cases, the structure is equally trivially preserved even if the factors are different. – GEdgar Mar 12 '13 at 17:23
@GEdgar Yes, thanks. I updated my question. Where can I read, what structure is preserved by taking the (Cartesian) product? (E.g. total order is not preserved, so there are caveats). – JSG Mar 12 '13 at 17:44
In category theory, there is a standard question of whether the category has products. And in a "concrete" category, whether the underlying set of a product is the cartesian product of the sets. – GEdgar Mar 12 '13 at 17:58

For algebraic structures (e.g. vector spaces, complete lattices) and relational structures (e.g. posets), or more generally models of a cartesian theory $\mathbb{T}$ (e.g. categories), the product of an arbitrary family of $\mathbb{T}$-models again a $\mathbb{T}$-model with componentwise operations and relations. This is easiest to understand in the case where $\mathbb{T}$ is an algebraic theory, i.e. a first-order theory over a signature with no relations, whose axioms are quantifier-free equations: it is clear that an equation involving componentwise operations is true if and only if it is true componentwise.

However, metric spaces are not structures of this kind. The fact that the product of finitely many metric spaces is again a metric space in a natural way is not explainable by general nonsense – and indeed, one sign that something subtle is happening is the fact that the sup metric on the product of infinitely many metric spaces does not in general induce the product topology. (At any rate, the product metric is not an example of a "componentwise" structure.)

It's also worth mentioning that the set of structure-preserving maps between two structures of the same type is in general not a structure of that type under componentwise operations. For example, although it is true that the set of homomorphisms between two abelian groups is again an abelian group, it is not true in general that the set of homomorphisms between two rings is again a ring. In fact, for one-sorted algebraic theories $\mathbb{T}$, one can show that a necessary (but not sufficient) condition for the set of $\mathbb{T}$-homomorphisms to be a $\mathbb{T}$-model is that $\mathbb{T}$ has at most one constant symbol (up to provable equality).

share|cite|improve this answer

The slice category $\mathbf{C}/C$ is the category of arrows with codomain $C$ and "commutative triangles" between them. In $\mathbf{Set}$, one can think to $\mathbf{Set}/X$ as the category of $X$-valued functions.

A thoerem of Freyd (Theorem 2.1 of Toposes, Triples and Theories) states that the property of $\mathbf{C}$ being a topos (which $\mathbf{Set}$ enjoys) is inherited by $\mathbf{C}/C$. So the slice category has a lot of structure, notably finite (co-)completeness, existence of exponentials, suboject classifier as well as the Heyting lattice structure of the subobjects of the terminal object. For a richly exemplified book on topoi, see Goldblatt's book.

If your space of interest is a topos, using the previous result may translates some of the properties you are interested in.

share|cite|improve this answer
Merely knowing that $\textbf{Set}_{/ X}$ is a topos is not very useful though. It's more important to note that the étale geometric morphism $\textbf{Set}_{/ X} \to \textbf{Set}$ has a logical inverse image functor. – Zhen Lin Mar 13 '13 at 10:00

Let me reiterate with Zhen Lin wrote, but in more classical language:

Proposition. Let $\sigma$ denote a (possibly infinitary) algebraic signature and suppose $X$ is an $I$-indexed family of $\sigma$-structures. Suppose $\eta$ is a (possibly infinitary) identity in the language of $\sigma.$ Then if every $X_i$ satisfies $\eta$, then $\prod_{i \in I} X_i$ also satisfies $\eta$.

Now define that $X^I = \prod_{i \in I} X$ for any $\sigma$-structure $X$. Then we have:

Corollary. Let $\sigma$ denote a (possibly infinitary) algebraic signature and suppose $X$ is a $\sigma$-structure. Then every (possibly infinitary) identity in the language of $\sigma$ that is satisfied by $X$ is also satisfied by $X^I$.

Excepting metric spaces, this covers all the examples you mention--even complete lattices and other models of infinitary signatures.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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