The center of a category $C$ is defined to be $\text{Z}(C) := \text{End}(1_C)$; here $1_C$ is the identity functor $C \to C$. See this question for an important application of the center. Now this n-lab entry states that $\text{Z}$ is functorial with respect to equivalences. But I think that we have to be careful here:

If $F: C \to D$ is an equivalence of categories, then we have to choose an inverse equivalence $G : D \to C$ and an isomorphism $e : 1_D \cong FG$ in order to define $\text{Z}(F) : \text{Z}(C) \to \text{Z}(D)$, namely by $s \mapsto e^{-1}(FsG) e$.

Thus in order to get a functor, we have to take the following category: Objects are categories, and a morphism $C \to D$ is a triple $(F,G,e)$, consisting of functors $F : C \to D, G : D \to C$ and an isomorphism $e : 1_D \cong FG$. It is clear how to define the composition and the identity. Then $\text{Z}$ is a functor on this category to the category of commutative monoids.

a) Is this correct so far? Has this already written down somewhere in the literature?

I'm wondering myself why this is not remarked in the context of the Reconstruction Theorem of Rosenberg, since there we have to use $C \cong D \Rightarrow \text{Z}(C) \cong \text{Z}(D)$ in order to reconstruct the structure sheaf.

b) Are my remarks superfluous since category theorists usually regard an equivalence not as a mere functor, but rather as an adjunction, whose functors are equivalences? Namely, this data includes the data required to define the functor $\text{Z}$ above.

By the way, the decategorified version of this is the center of a monoid (especially of a group). If $f : M \to N$ is a monoid isomorphism, then $f^{-1}$ is unique and $f f^{-1}$ is equal to $1_N$, thus we don't have to make any choices.

  • $\begingroup$ [deleted earlier comment since it was based on me misreading what Martin had written] $\endgroup$ – Yemon Choi Apr 10 '11 at 9:15
  • $\begingroup$ I think your (b) is the right one $\endgroup$ – Buschi Sergio Apr 10 '11 at 9:20
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    $\begingroup$ you also get a functor Z(F):Z(C)-> Z(D) for each fully faithful F:C-> D. $\endgroup$ – Steve Lack Apr 29 '11 at 11:40
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    $\begingroup$ sorry, I meant to say for each fully faithful F:D->C. $\endgroup$ – Steve Lack Apr 29 '11 at 21:41

You can make $Z(F)(s)$ one object at a time by just choosing, for each object $d$, an object $c$ and an iso $e:F(c)\to d$, and conjugating $s$ by $e$. This is independent of choices.

  • $\begingroup$ Why is this independent of choices? $\endgroup$ – Martin Brandenburg Apr 10 '11 at 19:06
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    $\begingroup$ If $(c,e:F(c)\to d)$ and $(c',e':F(c')\to d)$ are two such choices, then $e=e'\circ F(t)$ for some isomorphism $t:c\to c'$. Now use naturality of $s$. $\endgroup$ – Tom Goodwillie Apr 10 '11 at 20:04
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    $\begingroup$ ah, thanks! so the center is functorial with resp. to equivalences :) $\endgroup$ – Martin Brandenburg Apr 11 '11 at 8:37

It should be mentined that any localisation functor $F: C \to D$ induces a morphism $Z(F): Z(C) \to Z(D)$. A reference for this is Gabriel's thesis "Des categories abeliennes" (p. 446).

  • $\begingroup$ Yes. This is also Lemma 4.3 in my write-up arXiv:1310.5978 $\endgroup$ – Martin Brandenburg May 8 '14 at 7:02
  • $\begingroup$ Welcome to MathOverflow! $\endgroup$ – David White May 8 '14 at 14:38

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