In sufficiently nice categories, we can define an exponential object $B^A$ for all objects $A$ and $B$. Now I tend to think of $B^A$ as an internalization of $\mathrm{Hom}(A,B)$ to the category. This begs the question: given an object $A$ living in a sufficiently nice category, is there a way to define a new object $A!$ that somehow "internalizes" the set of all isomorphisms $A \rightarrow A$?


Yes if the ambient category $\mathcal{C}$ has pullbacks (and, of course, is cartesian closed). I will denote $B^A = \underline{\mathrm{Hom}}(A,B)$ and construct a subobject $\underline{\mathrm{Isom}}(A,B)$ as follows: Consider the morphisms $$\underline{\mathrm{Hom}}(A,B) \times \underline{\mathrm{Hom}}(B,A) \to \underline{\mathrm{Hom}}(A,A) \times \underline{\mathrm{Hom}}(B,B) \longleftarrow \star$$ given by $(f,g) \mapsto (gf,fg)$ and $(id_A,id_B) \leftarrow \star$ (if the notation is not clear, ask). Let $\underline{\mathrm{Isom}}(A,B)$ be their pullback. The composition $\underline{\mathrm{Isom}}(A,B) \to \underline{\mathrm{Hom}}(A,B) \times \underline{\mathrm{Hom}}(B,A) \to \underline{\mathrm{Hom}}(A,B)$ is a monomorphism, which exhibits the subobject of isomorphisms from $A$ to $B$.

For $A=B$ I wouldn't use $A!$ as a notation, but rather $\underline{\mathrm{Aut}}(A)$.

A typical example is $\mathcal{C}=\mathsf{Sh}(X)$, the category of sheaves on a space $X$. If $A,B$ are sheaves on $X$, the isomorphism sheaf $\underline{\mathrm{Isom}}(A,B)$ is the subsheaf of the usual homomorphism sheaf $\underline{\mathrm{Hom}}(A,B)$ whose sections on $U \to X$ are the isomorphisms $A|_U \to B|_U$.

  • $\begingroup$ I like your answer but I have to object that to my intuition the factorial is something different, maybe more naive: $n!=$ "what you get when you multiply $2\cdot 3\dots (n-1)\cdot n$". Your procedure obviously gives rise to the categorification of the automorphism group of a finite set, but I think we've lost the "combinatorial" side of the story. Multiplication is of course the decategorification of a monoidal product, so I would have taken a different path, albeit extremely more naive. $\endgroup$ – Fosco Jan 9 '14 at 22:30
  • $\begingroup$ For example (absolutely without any desire to superimpose my opinion to yours) I think that a natural definition for $[n]^{\times }!$ in the category of simplicial sets would be the cartesian product of all representables $[n]\times [n-1]\times\dots \times [2]\times [1]\times [0]$; or even more simply in the category $\Delta$ of simplices, $[n]^\oplus!$ would be... $\endgroup$ – Fosco Jan 9 '14 at 22:33
  • $\begingroup$ Have you read the question at all? It was about an internalization of the set of automorphisms of an object. Also observe that your product formula doesn't make sense in general categories (in fact not even for the category of sets, when the set is infinite). $\endgroup$ – Martin Brandenburg Jan 9 '14 at 22:58
  • $\begingroup$ Of course I read it! My point is that I wouldn't call such a thing "factorial"; but nevermind, it's not so important. I didn't want to give an answer, only to point out that to my intuition the factorial is an arithmetic notion, whose categorification is linked to the categorification of operations between natural numbers, much more than to the notion of automorphism. I'm sorry if I intruded! $\endgroup$ – Fosco Jan 9 '14 at 23:05
  • $\begingroup$ @tetrapharmakon, actually I called it factorial; Martin advocated calling it $\underline{\mathrm{Aut}}(A)$. Anyhow, the intent of the question was not to capture the combinatorial aspects of factorial, so much as about internalizing the group of isomorphisms of an object. $\endgroup$ – goblin Jan 10 '14 at 4:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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