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Are there any useful characterisation of categories whose auto-equivalences are all naturally isomorphic to the identity? For example, I read in this thread, What are the auto-equivalences of the category of groups? , that the category of groups is one such category. Is there some nice general property that I can use to check whether or not a category admits of auto-equivalences that are not naturally isomorphic to the identity? If not, it'd be useful just to find some more interesting examples of categories with this property.

More specifically, what examples (if any) are there of auto-equivalences that send every object to an isomorphic object but are not naturally isomorphic to the identity? For instance, I know that any auto-equivalence of SETS must send every object to an isomorphic object, but is it true that any such equivalence is naturally isomorphic to the identity? If so, is this a general property of toposes or just a special case?

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    $\begingroup$ If your category is a group G, then autoequivalences up to natural isomorphism are in bijection with the group of outer automorphisms Out(G). $\endgroup$ – Phil Tosteson Jun 26 '17 at 17:08
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    $\begingroup$ In particular, an outer automorphism of a group G induces an automorphism of the corresponding one-object groupoid BG that is not naturally isomorphic to the identity, but which sends the unique object to an isomorphic one (namely itself). $\endgroup$ – Mike Shulman Jun 27 '17 at 11:12
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    $\begingroup$ If $C$ is the category of finite sets with $6$ objects with bijections as morphisms, then $C$ has an auto-equivalence which takes a six-element set $X$ to its set of pentads (a "pentad" being a partition of the complete graph with vertex set $X$ into $5$ so that no two edges sharing a vertex are in the same class), which also has $6$ elements and is functorial in $X$. This auto-equivalence is not isomorphic to the identity. Conversely, for $n\neq 6$, the category of $n$-element sets and bijections has no non-trivial auto-equivalence. $\endgroup$ – Gro-Tsen Jun 27 '17 at 22:27
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Since you mention toposes, it may be worth pointing out a simple example of a topos that admits a nonidentity automorphism: $\mathrm{Set}\times\mathrm{Set}$ has an automorphism that swaps the two copies.

On the other hand, the proof that every automorphism of $\mathrm{Set}$ is the identity can be internalized to show that every indexed automorphism of any topos $\mathcal{E}$ is the identity. Here an "indexed automorphism" consists of automorphisms $\mathcal{E}/X \simeq \mathcal{E}/X$ for all $X\in\mathcal{E}$ that commute appropriately with the pullback functors $f^*:\mathcal{E}/Y \to \mathcal{E}/X$. Informally, since every topos "thinks that it is the category of sets", it also "thinks that every automorphism of itself is the identity". The argument is basically the same: in the world of $\mathcal{E}$-indexed categories, $\mathcal{E}$ itself is the free cocompletion of 1.

Similarly, if $\mathcal{V}$ is a complete and cocomplete closed symmetric monoidal category, then every $\mathcal{V}$-enriched automorphism of $\mathcal{V}$ is the identity. The proof is again the same: in the world of $\mathcal{V}$-enriched categories, $\mathcal{V}$ itself is the free cocompletion of 1.

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(This is essentially a more elementary rephrasing of what’s in Fosco Loregian’s answer.)

For instance, I know that any auto-equivalence of SETS must send every object to an isomorphic object, but is it true that any such equivalence is naturally isomorphic to the identity?

What is the argument by which you know this fact? The standard argument I know is — roughly stated — as follows:

  • any equivalence preserves coproducts and terminal objects
  • any set $X$ can be written as a coproduct $\coprod_{x:X} 1$
  • so any auto-equivalence $F$ sends $X$ to some $X$-indexed coproduct of copies of a terminal object, i.e. to some set isomorphic to $X$.

But if you look at this argument more carefully, it doesn’t just show that $FX$ is isomorphic to $X$; it gives you a specific isomorphism $X \to FX$, and this isomorphism is natural. Specifically, it’s the map which sends each $x \in X$ to $F(\ulcorner\! x\! \urcorner)(\ast) \in FX$, where $\ulcorner\! x\! \urcorner : 1 \to X$ picks out $x$, so $F(\ulcorner\! x\! \urcorner) : F1 \to FX$, and $\ast$ is the unique element of $F1$. This map $X \to FX$ is defined for any endofunctor $F$ of Set that preserves the terminal object, and it’s easy to check it’s natural; and it’s an iso when $F$ also preserves coproducts.

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    $\begingroup$ Thanks! That's pretty much the argument I had in mind for proving $F(X)$ is always isomorphic to $X$, but for some reason I didn't think about the specific functor this argument gives you. $\endgroup$ – King Kong Jun 27 '17 at 11:17
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is it true that any such equivalence is naturally isomorphic to the identity?

Does it work this way?

The presence of the Yoneda embedding $y_A : A \to [A°,Set]$ entails that there is an equivalence of categories $$ \hom(A,{\cal C}) \cong \hom_!([A°,Set],{\cal C}) $$ where $\cal C$ is cocomplete and $\hom_!$ is the category of functors that commutes with colimits. In particular if $A=*$ and ${\cal C}=Set$, you get $$ Set \cong \hom(*,Set) \cong \hom_!(Set,Set) $$ with an equivalence under which $1_{Set}$ and the singleton correspond each other.

(edit) Let's prove this statement: the equivalence of categories is induced by postcomposition with the Yoneda embedding, or equivalently given by the rule $$ \Big(F : A \to Set\Big)\mapsto Lan_yF $$ where $Lan_y$ is left Kan extension along the Yoneda embedding. There is a colimit formula for this Yoneda embedding, i.e. $$ Lan_yF(P) \cong \varinjlim_{ya\to P} Fa $$ and when instantiated to the case $F : * \to Set$ gives that $Lan_yF\cong F(*)\times-$ (cartesian product): this means that $F$ selects the terminal object of $Set$ iff $Lan_yF$ is the identity functor.

This means that $1_{Set}$ is terminal in $\hom_!(Set,Set)$, which is pretty much what you need to conclude.

(edit) ...because you said that "any auto-equivalence of SETS must send every object to an isomorphic object": if $\sigma$ is such an equivalence, then there is an isomorphism $X \to \sigma X$, but then its inverse must be the unique component of a natural transformation $\sigma \to 1$. This is saying no more no less than $\sigma\cong 1$ in a unique way (i.e. the unique morphism $\sigma \to 1$).

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  • $\begingroup$ Thanks! This is very helpful. I roughly understand the argument via the free co-completion property of the Yoneda embedding. However, I'm still a little unclear on a couple of points. (1) Why do we know that in the equivalence between $Set$ and $hom_{!}(Set, Set)$, $1_{Set}$ corresponds to the singleton? (2) Once we know that $1_{Set}$ is terminal in $hom_{!}(Set, Set)$, this guarantees that there is exactly one natural transformation from any auto-equivalence to the identity. Why does this ensure that the auto-equivalence is naturally isomorphic to the identity? $\endgroup$ – King Kong Jun 27 '17 at 7:58
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    $\begingroup$ I edited my answer :-) $\endgroup$ – Fosco Loregian Jun 27 '17 at 9:07

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