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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?

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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
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@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

2 Answers 2

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...

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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 $f^{-1} U \subseteq Y$.)

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+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

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