Definition. The centre of a category $\mathcal{C}$ is the set $\mathrm{Z}(\mathcal{C})$ defined by \begin{align*} \mathrm{Z}(\mathcal{C}) &\mathbin{\overset{\mathrm{def}}{=}} \int_{A\in\mathcal{C}}\mathrm{Hom}_{\mathcal{C}}(A,A)\\ &\cong \mathrm{Nat}(\mathrm{id}_{\mathcal{C}},\mathrm{id}_{\mathcal{C}}), \end{align*} while the trace of $\mathcal{C}$ is the set $\mathrm{Tr}(\mathcal{C})$ defined by \begin{align*} \mathrm{Tr}(\mathcal{C}) &\mathbin{\overset{\mathrm{def}}{=}} \int^{A\in\mathcal{C}}\mathrm{Hom}_{\mathcal{C}}(A,A)\\ &\cong \mathrm{End}(\mathcal{C})/\mathord{\sim}, \end{align*} where $\mathrm{End}(\mathcal{C})$ is the set of all endomorphisms of $\mathcal{C}$ and $\mathord{\sim}$ is the equivalence relation on $\mathrm{End}(\mathcal{C})$ generated by $g\circ f\sim f\circ g$. Here, the quotient map $$\mathrm{tr}\colon\mathrm{End}(\mathcal{C})\to\mathrm{Tr}(\mathcal{C})$$ is called the trace map.

Monoid Structure. The center of $\mathcal{C}$ comes with a natural monoid structure given by composition of natural transformations, and every monoidal category structure on $\mathcal{C}$ descends to a monoid structure on $\mathrm{Z}(\mathcal{C})$ (making it a commutative monoid by Eckmann–Hilton) and $\mathrm{Tr}(\mathcal{C})$.

Example. Take a group $G$ and view it as a one-object category $\mathrm{B}G$. Then, we can show that the centre of $\mathrm{B}G$ is the usual group-theoretic centre $\mathrm{Z}(G)$ of $G$, while the trace of $\mathrm{B}G$ is the set of conjugacy classes of $G$.

Question. What is the centre and trace of the simplex category $\Delta$?

What about the centre and trace of the augmented simplex category $\Delta_+$, which comes with a monoidal structure $\oplus$, thus making its centre a commutative monoid and its trace a monoid?

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    $\begingroup$ My guess is the center is trivial. Just look at morphisms from the one point total order. I would have to think about trace $\endgroup$ Apr 20 at 17:52
  • $\begingroup$ Dumb question: but is this definition of "center" a special case of the "center of an adjunction", for the identity adjoint equivalence of the category? ncatlab.org/nlab/show/fixed+point+of+an+adjunction $\endgroup$ Apr 21 at 20:12
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    $\begingroup$ @hasManyStupidQuestions Nope, the center of $\mathrm{id}_{\mathcal{C}}\dashv\mathrm{id}_{\mathcal{C}}$ is $\mathcal{C}$ itself, since we're looking at the full subcategory of $\mathcal{C}$ spanned by those objects $A$ of $\mathcal{C}$ such that the unit map $A\to\mathrm{id}_{\mathcal{C}}(\mathrm{id}_{\mathcal{C}}(A))$ of this adjunction is an isomorphism, which in this case is all of $\mathcal{C}$. $\endgroup$
    – Emily
    Apr 22 at 1:15

2 Answers 2


The center is trivial : as Benjamin said in the comments, an endomorphism of the identity is the identity on $\Delta^0$, and using the maps $\Delta^0\to \Delta^n$ you find that any endomorphism of the identity is the identity.

For the trace, you can use the following fact coming from unwinding the definition : if $f,g$ are composable in both orders in $C$, then $tr(fg) = tr(gf) \in Tr(C)$ where I let $tr$ of an endomorphism be the class it defines in $Tr(C)$.

Using this, for any endomorphism $f\in \Delta$, writing it as a surjection followed by an injection and inducting on size shows that every $tr(f)$ is $tr(id_S)$ for some $S\in \Delta$ (note that any automorphism is the identity). Furthermore, $f\mapsto $ the cardinality of the fixed point set of $f$ defines a morphism $Tr(\Delta)\to \mathbb N$ (this is not obvious, but a fun exercise in combinatorics) which distinguishes all the $tr(id_S), S\in \Delta$ so that $Tr(\Delta)\cong \mathbb N_{>0}$

  • $\begingroup$ Thank you so much, Maxime! I feel like this is probably easy, but do you see a quick way to show that $\#\mathrm{Im}(f\circ g)=\#\mathrm{Im}(g\circ f)$ for $f\colon[n]\to[m]$ and $g\colon[m]\to[n]$ morphisms of $\Delta$? I ask because $f\mapsto\#\mathrm{Im}(f)$ gives a map $\mathrm{Tr}(\Delta)\to\mathbb{N}_{\geq1}$ serving the same purpose of your fixed points map, but it also seems to be exactly the trace map $\mathrm{tr}\colon\mathrm{End}(\Delta)\to\mathrm{Tr}(\Delta)$, giving a quick description of which "conjugacy class" of $\Delta$ an endomorphism of it lives in. $\endgroup$
    – Emily
    Apr 22 at 1:08
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    $\begingroup$ @Emily: No, the size of the image doesn’t determine an element of the trace. You can falsify the required cyclicity property already with maps $f : [2] \to [3]$, $g : [3 ] \to [2]$. $\endgroup$ Apr 22 at 8:54
  • $\begingroup$ @PeterLeFanuLumsdaine Thank you, Peter! $\endgroup$
    – Emily
    Apr 22 at 15:47

$\newcommand{\Fix}{\mathrm{Fix}}\newcommand{\N}{\mathbb{N}}\newcommand{\id}{\mathrm{id}}\newcommand{\End}{\mathrm{End}}\newcommand{\Tr}{\mathrm{Tr}}\newcommand{\tr}{\mathrm{tr}}$Here are some extra details and explicit computations for determining the trace $\Tr(\Delta)$ of $\Delta$ as well as the trace map $$\tr\colon\End(\Delta)\to\Tr(\Delta),$$ following Maxime's answer.

An Explicit Example

Perhaps the general way to proceed is best exemplified by an example: consider the map $f\colon [4]\to[4]$ given by

and let's denote it (and maps of $\Delta$ in general) by the cleaner notation

To compute $\tr(f)$, we can write this map as the composition of a surjection $s\colon [4]\twoheadrightarrow[2]$ with an injection $i\colon[2]\hookrightarrow[4]$:

Now, since we have $\tr(f\circ g)=\tr(g\circ f)$ for any two endomorphisms $f$ and $g$ of $\Delta$ such that $f\circ g$ and $g\circ f$ are both defined, we see that $\tr(f)=\tr(i\circ s)=\tr(s\circ i)$, where $s\circ i$ is the map

which is the identity map $\id_{[2]}\colon[2]\to[2]$ of $[2]$ in $\Delta$.

The General Case

Claim 1. Let $f\colon[n]\to[n]$ be an arbitrary endomorphism of $\Delta$. We have $\tr(f)=\tr\big(\id_{[k]}\big)$ for some $k\leq n$.

Proof. If $f=\id_{[n]}$, then we are done. Otherwise, we can again decompose it into a surjection $s\colon[n]\twoheadrightarrow[k]$ followed by an injection $i\colon[k]\hookrightarrow[n]$, where $k=\#\mathrm{Im}(f)$ will be strictly less than $n$. By the property $\tr(f\circ g)=\tr(g\circ f)$ of the trace map, we have \begin{align*} \tr(f) &= \tr(i\circ s)\\ &= \tr(s\circ i), \end{align*} with $s\circ i\colon[k]\to[k]$ again an endomorphism of $\Delta$. Now, if $s\circ i=\id_{[k]}$, we are done. Otherwise, we repeat this process again.

Since $n$ is finite, it follows that repeating this process will give us a natural number $n_0\leq n$ such that $\tr(f)=\tr\big(\id_{[n_0]}\big)$ in finitely many steps.

Next, we aim to show the following:

Claim 2. If $n\neq m$, then $\tr\big(\id_{[n]}\big)\neq\tr\big(\id_{[m]}\big)$.

To this end, we want we want to construct a map $$\phi\colon\mathrm{End}(\Delta)\to\N_{\geq1}$$ satisfying the following conditions:

  1. If $n\neq m$, then $\phi\big(\id_{[n]}\big)\neq\phi\big(\id_{[m]}\big)$.
  2. We have $\phi(f\circ g)=\phi(g\circ f)$, so that $\phi$ descends to a map $\widetilde{\phi}\colon\mathrm{Tr}(\Delta)\to\N_{\geq1}$.

From the example of $f\colon[4]\to[4]$ above and similar ones, we might guess that the map $\sigma\mapsto\#\mathrm{Im}(\sigma)$ sending an endomorphism of $\Delta$ to the cardinality of its image would be our desired map $\phi$.

However, as Peter pointed out, the map $\sigma\mapsto\#\mathrm{Im}(\sigma)$ fails to satisfy Item (2) above, as the following example shows:

Consider the maps $f\colon[3]\to[2]$ and $g\colon[2]\to[3]$ given by

Then the maps $g\circ f\colon[3]\to[3]$ and $f\circ g\colon[2]\to[2]$ are given by

and we have $\#\mathrm{Im}(g\circ f)=2$ while $\#\mathrm{Im}(f\circ g)=1$.

The map sending $\sigma\colon[n]\to[n]$ to the cardinality $\#\Fix(\sigma)$ of its set of fixed points does fill both criteria:

  1. If $n\neq m$, then $\#\Fix(\id_{[n]})=n+1\neq m+1=\#\Fix(\id_{[m]})$.

  2. We have $\#\Fix(g\circ f)=\#\Fix(f\circ g)$, as the maps \begin{align*} f|_{\Fix(g\circ f)} &\colon \Fix(g\circ f) \to \Fix(f\circ g),\\ g|_{\Fix(f\circ g)} &\colon \Fix(f\circ g) \to \Fix(g\circ f) \end{align*} set up a bijection between $\Fix(g\circ f)$ and $\Fix(f\circ g)$, as:

    a) If $x$ is a fixed point of $g\circ f$, i.e. $g(f(x))=x$, then $f(x)$ is a fixed point of $f\circ g$, since $f(g(f(x))=f(x)$.

    b) Similarly, if $y$ is a fixed point of $f\circ g$, then $g(y)$ is a fixed point of $g\circ f$.

    c) For each $x\in\Fix(g\circ f)$, we have \begin{align*} [g|_{\Fix(f\circ g)}\circ f|_{\Fix(g\circ f)}](x) &= g(f(x))\\ &= x\\ &= [\id_{\Fix(g\circ f)}](x). \end{align*} so $g|_{\Fix(f\circ g)}\circ f|_{\Fix(g\circ f)}=\id_{\Fix(g\circ f)}$.

    d) Similarly, $f|_{\Fix(g\circ f)}\circ g|_{\Fix(f\circ g)}=\id_{\Fix(f\circ g)}$.

It follows that $\sigma\mapsto\#\Fix(\sigma)$ defines a map $\widetilde{\phi}\colon\Tr(\Delta)\to\mathbb{N}_{\geq1}$.

Finally, the result.


  1. For all $\tr(f)\in\Tr(\Delta)$, there exists some $n\in\mathbb{N}_{\geq1}$ such that $\tr(f)=\tr\big(\id_{[n]}\big)$ (Claim 1).
  2. If $n\neq m$, then $\tr\big(\id_{[n]}\big)\neq\tr\big(\id_{[m]}\big)$ (Claim 2).

it follows that $\widetilde{\phi}$ is a bijection:

  1. $\widetilde{\phi}$ Is Injective: Given $\tr(f)$ and $\tr(g)$ with $\widetilde{\phi}(\tr(f))=\widetilde{\phi}(\tr(g))$, there exist $n,m\in\mathbb{N}_{\geq1}$ such that $\tr(f)=\tr\big(\id_{[n]}\big)$ and $\tr(g)=\tr\big(\id_{[m]}\big)$ by Claim 1, but since $\widetilde{\phi}(\tr(f))=\widetilde{\phi}(\tr(g))$, we must have $n=m$ by Claim 2, so $\tr(f)=\tr(g)$.
  2. $\widetilde{\phi}$ Is Surjective: Given $n\in\N_{\geq1}$, we have $\widetilde{\phi}\big(\tr\big(\id_{[n-1]}\big)\big)=n$.

So $\Tr(\Delta)\cong\mathbb{N}_{\geq1}$. Lastly, to determine the trace map


note that we have $\phi=\tr\circ\widetilde{\phi}$ by the universal property of the quotient, and since $\widetilde{\phi}$ is a bijection, we have $\tr=\widetilde{\phi}{}^{-1}\circ\phi$, which is given by $\sigma\mapsto\tr\big(\id_{[\#\Fix(\sigma)-1]}\big)$.


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