In $Set$ the initial object is the empty set, and it has an unique morphism to each other object, namely $f=\emptyset$. However I find it difficult to think about the category ${Set}^{op}$, is there any morphism from $X \subset Obj(Set)$ to $\emptyset$ at all? Can an empty set be the codomain of some function?

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    $\begingroup$ The opposite category is the opposite category. There is no need to think about morphisms as actual functions with the specified domain and codomain; and indeed, you can't. $\endgroup$
    – Zhen Lin
    Oct 19, 2014 at 15:18
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    $\begingroup$ It's $\mathbf{teS}$. $\endgroup$
    – Asaf Karagila
    Oct 19, 2014 at 15:31
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    $\begingroup$ @AsafKaragila haha that's funny but not very helpful. $\endgroup$
    – Boyu Fang
    Oct 19, 2014 at 15:49
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    $\begingroup$ I think the question of that form is helpful to you: there is a concrete category which is isomorphic to the opposite category of $\mathbf{Set}$? $\endgroup$
    – Hanul Jeon
    Oct 19, 2014 at 16:39
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    $\begingroup$ No, there is no need at all for the morphisms of $\mathsf{Set}^\mathrm{op}$ to be some kind of concrete functions. What makes you say that ? $\endgroup$
    – Pece
    Oct 19, 2014 at 17:06

4 Answers 4


I only just saw this question. I'm a little surprised that none of the answers said concretely what $Set^{op}$ is. It's the category of complete atomic Boolean algebras, or equivalently complete Boolean algebras of the form $P(X)$ where $X$ is a set. The morphisms are morphisms of complete Boolean algebras (so functions that preserve arbitrary meets and negation, whence preserving the Boolean operations as well).

Note that a morphism of complete atomic Boolean algebras $\phi: P(X) \to P(Y)$ is uniquely determined by what it does to the atoms of $P(X)$. The atoms are of course singletons of $X$; if we denote $\phi(\{x\}) = S_x \subseteq Y$, then we can define a corresponding function $f: Y \to X$ to be the unique one such that $f^{-1}(x) = S_x$. Note that $S_x$ and $S_{x'}$ are disjoint if $x \neq x'$, because $\emptyset = \phi(\emptyset) = \phi(\{x\} \cap \{x'\}) = \phi(\{x\}) \cap \phi(\{x'\}) = S_x \cap S_{x'}$. This shows that $f$ is well-defined. This should give a clear idea why we obtain the opposite category.

The concrete manifestation of $X \to \emptyset$ in $Set^{op}$ is the unique map $P(X) \to P(\emptyset)$, where the codomain has just one element.


In abstract category theory morphisms are not required to be functions: we are given an arbitrary class, regarding its elements as arrows, equipped with domain and codomain information and an associative operation.

If we want, we can specify so that for a funtion $f:A\to B$, let ${\rm dom\,}f:=B$ and ${\rm cod\,}f:=A$ and the operation is defined as reversed composition of functions -- this is the category $Set^{op}$.
The same entity, $f$, as a function, plays a role of an arrow $f:A\to B$ in $Set$, but at the same time, itself is an arrow $B\to A$ in $Set^{op}$.

Anyway, the contravariant powerset functor establishes an equivalence between the opposite of category of finite sets and the category $BA_{fin}$ of finite Boolean algebras, and hence $$Set^{op} \simeq Pro(BA_{fin})$$ where $Pro$ denotes the pro-completion: i.e. taking all filtered limits formally.

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    $\begingroup$ By Stone duality there is an equivalence of categories between (locally compact totally disconnected Hausdorff spaces)$^{op}$ and (non-unital boolean algebras). We may restrict to discrete spaces on the left hand side and get (Set)$^{op}$. What do we get on the right hand side? Perhaps complete boolean algebras? $\endgroup$ Oct 22, 2014 at 13:05
  • $\begingroup$ Is $f^{\textbf{op}}$ of $f:A \rightarrow B$ not strictly a 'function' in the usual sense? $\endgroup$
    – John Mars
    Mar 5, 2021 at 17:08

This is a pretty old question, so I'm a bit suprised that nobody mentioned the most simple answer.

Since functions are the right-unique, left-total binary relations, they reverses are the left-unique, right-total ones. So, $\mathsf{Set}^\mathrm{op}$ is the category with sets as objects and left-unique, right-total binary relations as morhisms.

Regarding relations as subsets of the Cartesian product, the $X\to \emptyset$ morphism is the empty set. This is really a morphism in $\mathsf{Set}^\mathrm{op}$ because it is a left-unique and right total binary relation.

  • $\begingroup$ Other terminology I've seen is that functions are "cosurjective coinjective relations", so morphisms in $\mathsf{Set}^{\mathrm{op}}$ are "surjective injective relations." $\endgroup$ Aug 22, 2021 at 7:58

There is exactly one morphism from $X$ to $\emptyset$, namely the unique function from $\emptyset$ to $X$.

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    $\begingroup$ In other words, recall that a morphism from $A$ to $B$ in $\textbf{Set}^{op}$ is nothing else but a function $B\to A$. $\endgroup$ Oct 22, 2014 at 13:06

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