# Is it possible to make an autoequivalence of categories an automorphism?

In studying triangulated categories, some authors require the shift functor $T: \mathcal D \rightarrow \mathcal D$ to be an autoequivalence, whereas others require it to be an automorphism (i.e. strictly invertible). Unfortunately, I couldn't find any reference which clarifies whether the two requirements are actually equivalent or not. Indeed, I think they are, so, abstracting a little, here is my question: given an autoequivalence $T: \mathcal C \rightarrow \mathcal C$ of an arbitrary category $\mathcal C$, is it possible to find a functor $T': \mathcal C \rightarrow \mathcal C$ which is isomorphic to $T$ and is also an automorphism of $\mathcal C$?

More generally, one could ask if a similar result is true for functors between categories whose object sets are of the same cardinality...

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For the second question, no: the categories need not be isomorphic. Corresponding isomorphism classes might have different cardinalities. –  Tom Goodwillie Mar 22 '12 at 11:31
even with categories whose object sets have same cardinalities I think you can cook up artificial counterexamples (by playing with discrete categories for example). –  Yosemite Sam Mar 22 '12 at 12:18
martin's answer is nice. when I read Kashiwara-Schapira's treatment of derived categories (in the cats and shvs book) I was upset by the shift being an equivalence. If you think about it though it makes more (categorical) sense to set up the theory this way and it practice you never really care. –  Yosemite Sam Mar 22 '12 at 12:19
Converse to my earlier comment: Of course if $T:C\to D$ is an equivalence, and if for every object $x$ there is a bijection from the iso class of $x$ to that of $Tx$, then you can define $T'$ on objects by using these bijections and also choose an isomorphism $TX\to T'x$ for each $x$. Requiring this isomorphism $T\to T'$ to be natural forces your choice of how to specify $T'$ on morphisms. –  Tom Goodwillie Mar 22 '12 at 13:36

In practice one may always assume that such a shift is an automorphism instead of an auto-equivalence. But for that one also has to modify the category with an equivalence (but this is OK for applications):

Let $F : C \to C$ be an equivalence of categories. Define the following category $C'$: Objects are sequences of objects $(X_n)_{n \in \mathbb{Z}}$ with $X_n \in C$, together with isomorphisms $\alpha_n : F(X_n) \cong X_{n+1}$. It is clear how to define the morphisms. There is a canonical automorphism $F' : C' \to C'$, which is just a shift $(X_n,\alpha_n)_n \mapsto (X_{n+1},\alpha_{n+1})_n$. Besides the evaluation at $0$ yields a functor $C' \to C$. It is easy to check that this is an equivalence of categories which makes the diagram

$$\begin{matrix} C' & \stackrel{F'}{\rightarrow} & C' \\\\ \downarrow && \downarrow \\\\ C & \stackrel{F}{\rightarrow} & C \end{matrix}$$

commute up to natural isomorphism of functors.

To put this into more global terms: Let $\mathcal{A}$ be the $2$-category of all categories equipped with an auto-equivalence, and $\mathcal{A}'$ the full $2$-subcategory of $\mathcal{A}$ consisting of those categories equipped with an automorphism. Then $\mathcal{A}' \hookrightarrow \mathcal{A}$ is an $2$-equivalence of $2$-categories.

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Thanks; I think I will use the "autoequivalence" definition for triangulated categories, in future: very better behaved, no technical drawbacks! –  Francesco Genovese Mar 22 '12 at 13:53
On the other hand, one might argue that the phrase "automorphism of categories" should mean "autoequivalence", since isomorphisms of (large) categories are rarely interesting or important. –  Mike Shulman Mar 22 '12 at 16:14
@Mike: I don't agree. The first definition of a category is just a graph with additional structure. It makes perfect sense to talk about the $1$-category of all (insert size bound...) categories. Now we can apply the usual definition of "automorphism" to that category. It is quite convenient to work with automorphisms of categories instead of auto-equivalences because you don't have to worry about the compatibility data. For a specific application see the paper "Noncommutative projective schemes" by Artin-Zhang, specifically Section 4. –  Martin Brandenburg Mar 23 '12 at 0:50
An important and interesting automorphism of some large category is the shift operator on chain complexes valued in an abelian category. // Similarly (but I'm sure you know all that, but perhaps this is also interesting for others ...) it is quite convenient to work with strict monoidal functors instead of strong monoidal functors. In fact, in practice no one wants to keep track of all these compatibility isomorphisms. There is a result (essentially with the same proof as above) that every strong monoidal functor is equivalent to a strict monoidal functor when we modify the domain category. –  Martin Brandenburg Mar 23 '12 at 0:54