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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.

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  • $\begingroup$ [deleted earlier comment since it was based on me misreading what Martin had written] $\endgroup$
    – Yemon Choi
    Commented Apr 10, 2011 at 9:15
  • $\begingroup$ I think your (b) is the right one $\endgroup$
    – Buschi Sergio
    Commented Apr 10, 2011 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
    Commented Apr 29, 2011 at 11:40
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    $\begingroup$ sorry, I meant to say for each fully faithful F:D->C. $\endgroup$
    – Steve Lack
    Commented Apr 29, 2011 at 21:41

2 Answers 2

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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.

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  • $\begingroup$ Why is this independent of choices? $\endgroup$
    – Martin Brandenburg
    Commented Apr 10, 2011 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
    Commented Apr 10, 2011 at 20:04
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    $\begingroup$ ah, thanks! so the center is functorial with resp. to equivalences :) $\endgroup$
    – Martin Brandenburg
    Commented Apr 11, 2011 at 8:37
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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).

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  • $\begingroup$ Yes. This is also Lemma 4.3 in my write-up arXiv:1310.5978 $\endgroup$
    – Martin Brandenburg
    Commented May 8, 2014 at 7:02
  • $\begingroup$ Welcome to MathOverflow! $\endgroup$
    – David White
    Commented May 8, 2014 at 14:38

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