Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

How to interpret the morphism $1 \to 0$ in ${\bf Set}^\mathrm{op}$, dual to $\bf Set$, with the standard meanings of the initial and terminal objects? Since the objects have the same interpretation in the dual, can ${\bf Set}^\mathrm{op}$ be interpreted (and if so uniquely?) as the category of sets and partial functions?

share|cite|improve this question
I don't think you can interpret $\bf Set^\mathrm{op}$ as the category of sets and partial functions. Consider the morphism of $\bf Set$ from ${\mathbb Z}\to\{0, 1\}$ which maps even numbers to 0 and odd numbers to 1. There is no way to understand this as a partial function $\{0, 1\}\to{\mathbb Z}$. – MJD May 3 '12 at 2:25
Understood. Since functions = total functional relations, it seems neither of those properties are preserved? Can Set* be considered the cat of sets and relations? (ps, thanks for the markup - where can I learn this code?) – alancalvitti May 3 '12 at 2:41
@alan: $\text{Set}^{op}$ embeds into the category of sets and relations, but is not the whole thing. – Qiaochu Yuan May 3 '12 at 3:10
@alan: the markup language is called $\TeX$. You can find many tutorials with a web search. – MJD May 3 '12 at 4:02
@alancalvitti you can see and copy/paste the Tex code used in questions by trying to Edit the question (at the end , of course, you Discard the Edit). I do not know how you look at the code used in comments. – magma Jul 27 '12 at 8:38
up vote 3 down vote accepted

If $f\colon A\to B$ in $\mathbf{Set}$, then the morphism in the opposite category $f^{op}\colon B\to A$ can be thought of as a multivalued partial function. That is, it is only defined on $\operatorname{im} f$ and the image of $f^{op}(b)$ is the set $f^{-1}(b)$.

If we try and recover exactly which conditions we need to place on our multivalued partial functions so that every one corresponds uniquely to a honest function of sets, we find that $\mathbf{Set}^{op}$ is equivalent to the category whose objects are sets and whose morphisms are surjective multivalued partial functions with the property that $f(a)$ and $f(b)$ are disjoint sets for $a\neq b$. If $f\colon B\to A$ is a morphism as defined above, the corresponding function $g\colon A\to B$ of sets is given by $g(a)=b$ where $b$ is the unique element such that $a\in f(b)$.

This is doubtfully a useful interpretation...

share|cite|improve this answer
It's useful (@ least to me) for understanding duality, thanks – alancalvitti May 4 '12 at 0:52

It is a remarkable fact that $\textbf{Set}^\textrm{op}$ is actually a completely concrete category: it is naturally equivalent to the category of complete atomic boolean algebras via the contravariant power set functor. Thus, an object $X$ in $\textbf{Set}^\textrm{op}$ secretly stands for its powerset $P X$, and a morphism $X \to Y$ in $\textbf{Set}^\textrm{op}$ is then a homomorphism of complete boolean algebras $P X \to P Y$. (More precisely, if $f : Y \to X$ is a map in $\textbf{Set}$, then the corresponding homomorphism $P f : P X \to P Y$ is the one that sends a subset $U \subseteq X$ to its preimage $f^{-1} U \subseteq Y$.)

share|cite|improve this answer
+1. Two Q's: do you concur w/ SL2's characterization of $\textbf{Set}^\textrm{op}$, and if you have time, do you mind editing your answer w/ additional detail on this interesting construction? – alancalvitti Feb 21 '13 at 18:31

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.