Segal's category $\Gamma$ is the skeleton of the category $\text{FinSet}_{\ast}$ of pointed finite sets. It is used to write down $\Gamma$-spaces, which are functors $\Gamma \to \text{Top}$ satisfying some conditions, and which model infinite loop spaces. I would like to be able to tell myself a story about this category which would explain in some sense why one might have come up with it as a candidate to be part of a delooping machine.

For comparison, here is the analogous story about $\text{FinSet}$: equipped with disjoint union, it is the free symmetric monoidal category on a commutative monoid. (I guess all of my stories are universal properties.) Since infinite loop spaces are in particular supposed to be like homotopy coherent commutative monoids I can see how one might have come up with $\text{FinSet}$ as a candidate to model infinite loop spaces, but not $\text{FinSet}_{\ast}$.

Riffing off of the above, it seems like $\text{FinSet}_{\ast}$, equipped with wedge sum, is the free symmetric monoidal category on something like a "copointed" commutative monoid; that is, a commutative monoid together with a map $\varepsilon : M \to 1$. The idea is that $\text{FinSet}_{\ast}$ can equivalently be thought of as the category of sets and partial functions, and throwing in a map $\varepsilon : M \to 1$ lets us model partial functions by using $\varepsilon$ to throw away the points at which a partial function isn't defined.

Why is this, and not $\text{FinSet}$, a reasonable category to use to model infinite loop spaces? (The inclusion of $\varepsilon : M \to 1$ is particularly strange because in any cartesian monoidal category, such as $\text{Top}$, it is unique because $1$ is the terminal object.) I guess $\varepsilon$ is needed to get an inclusion of $\Delta^{op}$ into $\Gamma$ so we can define the geometric realization of a $\Gamma$-space, but now I don't understand why there should be such an inclusion; the universal properties don't suggest it. The augmented simplex category $\Delta_a$ is the free monoidal category on a monoid, so the universal properties suggest instead a monoidal functor $\Delta_a \to \Gamma$.

  • 5
    $\begingroup$ Segal's Gamma is not FinSet*, but its opposite. Infinite loop $\endgroup$
    – Peter May
    Oct 9, 2013 at 1:40
  • 3
    $\begingroup$ Yes, and a Gamma-space is a functor from the opposite of Gamma to Top. Oddly, in the paper where Segal introduced all of this he never mentioned the interpretation of Gamma, the opposite of the category he used, is a skeleton of FinSet_*. $\endgroup$ Oct 9, 2013 at 2:16
  • 2
    $\begingroup$ That's the convention Segal uses, I guess? The nLab is somewhat inconsistent about which category ought to be called Gamma. $\endgroup$ Oct 9, 2013 at 3:58
  • 1
    $\begingroup$ I think Proposition 3.1.5 in Leinster's "Homotopy algebras for operads" is relevant. Namely, $\Gamma^{op}$ is the Kleisli category for the monad $1\times -$ on finite sets (and Fin is the prop for the commutative monoid operad, as you say). Similarly, $\Delta^{op}$ is the Kleisli category for the monad $1\times - \times 1$ on finite ordinals. This passage was studied more systematically by Barwick in "From operator categories to topological operads". $\endgroup$ Nov 17, 2017 at 18:00

4 Answers 4


Infinite loop spaces and spectra are intrinsically pointed, and the purpose of the basepoint is to build in basepoints, which give the units for the associated products. Let $T_*$ be the category of based objects in any cartesian monoidal category T. For an object $X$ of $T_*$, a covariant functor $X^*: F_* \longrightarrow T_*$ that sends the based set $n=\{0,1, \cdots, n\}$ with basepoint $0$ to $X^n$ is precisely a commutative monoid in $T$ with its unit element equal to the basepoint of $X$. Use of basepoints like this long precedes Segal. When $T$ is spaces, the James construction $JX$ is the free monoid on $X$ with unit the basepoint of $X$ and the infinite symmetric product $NX$ on $X$ is the free commutative monoid on $X$ with its unit element the basepoint of $X$, both suitably topologized.

In more detail, the morphisms of $F$ are generated under composition by injections, projections, and the based maps $\phi_n \colon n\to 1$ that send $i$ to $1$ for $1\leq i\leq n$. (Using the wedge sum, only $\phi_2$ need be added). The morphisms $\pi\colon m\to n$ such that $\phi^{-1}(i)$ has $0$ or $1$ element give a subcategory $\Pi$ of $F$, and the functor $X^*$ has underlying functor $\Pi\to T_*$ given by the injections (determined by the basepoint), projections, and permutations that are given by the assumption than $n \mapsto X^n$. The map $\phi_2$ gives a product $X\times X\longrightarrow X$ and, more generally, the $\phi_n \colon X^n\longrightarrow X$ give the unique $n$-fold product determined by $\phi_2$.

The point of infinite loop space theory is to build in the axioms of a commutative monoid up to "all higher coherence homotopies", and the genius of Segal was to see that the evident maps $\delta_i\colon n\longrightarrow 1$ that send $i$ to $1$ and $j$ to $0$ for $j\neq i$ can be used to build in these homotopies. Taking $T$ to be spaces, for a functor $Y\colon F_* \longrightarrow T_*$, we have based spaces $Y_n$, and the $\delta_i$ determine the Segal maps $\delta^n\colon Y_n \longrightarrow Y_1^n$. Requiring these maps to be homotopy equivalences for all $n\geq 0$ makes $Y_1$ a "commutative monoid up to all higher coherence homotopies", with canonical zigzag product $Y_1^2 \longleftarrow Y_2 \longrightarrow Y_1$ determined by $\delta^2$ and $\phi_2$.

I could go on forever about this. But maybe I'll just give the definition of the functor $K\colon \Delta^{op}\longrightarrow F_*$. I agree that its interpretation may not be obvious. Of course, $Kn = n$. For a map $\phi\colon n\longrightarrow m$ in $\Delta$, define $K\phi\colon m\longrightarrow n$ by sending $i$ to $j$ whenever $\phi(j-1) < i \leq \phi(j)$, where $1\leq j\leq n$, and sending $i$ to $0$ if there is no such $j$. Thus $$ (K\phi)^{-1}(j) = \{ i | \phi(j-1) < i \leq \phi(j)\} \ \ \text{for} \ \ 1\leq j\leq n. $$

Please excuse the lousy ad hoc notation (F= finite sets).

  • $\begingroup$ So this explanation confuses me because $\text{FinSet}$ already seems to incorporate information about basepoints: namely, the empty set $\emptyset$ is floating around in there already with a unique map to every other object, so if I require that a functor $F : \text{FinSet} \to \text{Top}$ (say) satisfy the property that $F(\emptyset)$ is a point then I already get basepoints on $F(n)$ for all $n$. $\endgroup$ Oct 9, 2013 at 4:09
  • $\begingroup$ I think what you're saying in the first paragraph amounts to saying that $\text{FinSet}_{\ast}$ is the free "pointed symmetric monoidal category" on a commutative monoid, where by "pointed symmetric monoidal category" I guess I mean a symmetric monoidal category such that the tensor unit is also a zero object. In particular based spaces and so forth are such categories. But again, I don't see why $\text{FinSet}$ hasn't already said everything that needs to be said about basepoints in this story. $\endgroup$ Oct 9, 2013 at 4:10
  • 3
    $\begingroup$ Qiaochu, it might help to focus more on the point of the definition than on the formalities. If you just use the category of finite sets, you don't have the projections that are used to define the Segal maps. Basepoints give targets to which one can send points (``project coordinates'') one wants to ignore. One wants the domain category to see all of the structure that is automatic in the example n \mapsto X^n. I think that is the key point. $\endgroup$
    – Peter May
    Oct 9, 2013 at 16:52

Just a remark about the functor $\Delta^{op}\to \Gamma$. One way to think about it is this:

An object of $\Delta$ is a nonempty ordered finite set $S$. To such a thing, associate another finite ordered set whose elements are all possible "cuts" in $S$: all ways of dividing $S$ into a left part and a right part, including the two extremes where either the left or the right part is empty. This establishes an equivalence of categories between

$\Delta=$(nonempty ordered finite sets, monotone maps)

and the opposite of

(ordered finite sets with two endpoints, monotone end-point-preserving maps).

Now map the latter to

(based finite sets, based maps)

by the functor that identifies the two endpoints.

  • $\begingroup$ Another remark about that functor. We may view it as a contravariant functor landing in the (skeletal) category of finite based sets. It is then nothing but the standard simplicial circle S^1 = \Delta[1]/\partial \Delta[1]. $\endgroup$
    – Peter May
    Apr 15, 2016 at 23:59

Not an answer really, just a few random remarks.

Segal's category $\Gamma$ can be described as the one with

  • objects: finite sets $S$,
  • morphisms $S\to T$: pairs $(T_0\subseteq T,f\colon T_0\to S)$.

Composition is like composition of spans. This is basically the definition Segal gives (Segal's description is a bit more obscure than this.)

As Peter comments, $\mathrm{FinSet}_*$ is essentially opposite category of Segal's $\Gamma$, and $\Gamma$-spaces are presheaves on $\Gamma$. The convention seems to have shifted to write $\Gamma$ for $\mathrm{FinSet}_*$ itself; I think that turns out to be a bad choice.

Any abelian group $A$ gives you a presheaf $F$ on $\Gamma$ by: $F(S)=\{\text{functions $S\to A$}\}$, with $F(T)\to F(S)$ defined by "restrict to $T_0$, then integrate along $f$".

The functor $\Delta\to \Gamma$ is a bit funny to describe, and it isn't actually an inclusion of a subcategory (there's only one map $\{\}\to \{1\}$ in $\Gamma$, but two maps $[0]\to [1]$ in $\Delta$.)

  • 2
    $\begingroup$ Charles, you beat me to it. I agree about your ``bad choice'' comment; I've been using script F for finite pointed sets for decades. I wrote down the funny non-inclusion you mention. $\endgroup$
    – Peter May
    Oct 9, 2013 at 2:56

If we can agree that it is natural to use $\Delta$ to encode categories and their homotopical generalisations, then I don't think it is farfetched to view $\Gamma$ as the analogous gadget to encode commutative monoids and their homotopical generalisations.
So let's think a little about $\Delta$ first. The category $\Delta$ has objects in bijection to $\mathbb{N}$, so that a presheaf $C$ on $\Delta$ is a graded set; for every $n \in \mathbb{N}$ we want to think of $C(n)$ as a set of chains of composable morphisms of length $n$. The morphisms in $\Delta$ are supposed to encode things like composition and associativity; so for example there should be some morphism $[1] \to [2]$ which gives composition of pairs of morphisms. But how do we know that we have all necessary morphisms in $\Delta$? For every $n \in \mathbb{N}$ denote by $\Delta_n$ the free category generated by a graph consisting of a chain of $n$ directed arrows. For a small category $\mathscr{C}$ the set of chains of morphisms of length $n$ is canonically isomorphic to $\mathbf{Cat}(\Delta_n, \mathscr{C})$. We could now hope that there is a functor $\Delta \to \mathbf{Cat}, \; [n] \mapsto \Delta_n$ which induces the fully faithful functor $\mathbf{Cat} \hookrightarrow \widehat{\Delta}, \; \mathscr{C} \mapsto ([n] \mapsto \mathbf{Cat}(\Delta_n,\mathscr{C}))$. Indeed, if we choose this functor to be fully faithful, which completely determines the structure of $\Delta$, then we obtain such a functor $\mathbf{Cat} \hookrightarrow \widehat{\Delta}$, the classical nerve functor.
If we play the same game with monoids we obtain the $\Gamma$ category. If we view any commutative monoid as a category, then we see that we might again suppose that $\Gamma$ has objects in bijection to $\mathbb{N}$; for any number $n \in \mathbb{N}$ we call the corresponding object $(n)$. Now a commutative monoid has exactly one object, so without knowing anything else, we may already assume that $(0)$ gets mapped to $\{*\}$. Like for $\Delta$, let's try to view $\Gamma$ as a subcategory of $\mathbf{AbMon}$, the category of commutative monoids, in order to again obtain a nerve functor. The objects have to be $\mathbb{N} \oplus \underbrace{\cdots}_{n \times} \oplus \mathbb{N}$ for all $n \in \mathbb{N}$. The morphisms of $\Gamma$ are then completely determined by where the generators of $\mathbb{N} \oplus \underbrace{\cdots}_{n \times} \oplus \mathbb{N}$ are sent. To get an appropriate subcategory of $\mathbf{AbMon}$ we will only consider those morphisms which send generators to elements, which themselves are sums of distinct generators (this will be important below). If we label and keep track of only the generators, we rediscover Segal's original description of $\Gamma$. The induced nerve functor is again fully faithful, and it is now straightforward how to view any commutative monoid as a $\Gamma$-set. Furthermore, we note that there is a unique morphism $(n) \to (0)$ for all $n \in \mathbb{N}$, so that any presheaf on $\Gamma$ taking $(0)$ to $\{ * \}$ factors through the category of pointed sets $\mathbf{Set}_*$, and we might as well consider contravariant functors $\Gamma^{\mathrm{op}} \to \mathbf{Set}_*$.
Finally, let us show that there is a canonical functor $\Delta \to \Gamma$, so that every $\Gamma$-set has an underlying simplicial set. This is simple: by again viewing any commutative monoid as a category, we simply send the generators of $\Delta_n$ to the generators of $\mathbb{N} \oplus \underbrace{\cdots}_{n \times} \oplus \mathbb{N}$ for every $n \in \mathbb{N}$. In this last step we are implicitly using that we have labelled the generators of $\mathbb{N} \oplus \underbrace{\cdots}_{n \times} \oplus \mathbb{N}$, but this is not a problem: Let us denote the nerve functor $\mathbf{Cat} \hookrightarrow \widehat{\Delta}$ by $N$, and the nerve functor $\mathbf{AbMon} \hookrightarrow \mathbf{Cat}(\Gamma^{\mathrm{op}}, \mathbf{Set}_*)$ by $N_\Gamma$. Then for any monoid $M$, any functor $\Delta \to \Gamma$ induced by any labelling will take $N_\Gamma(M)$ to $N(M)$ (viewing $\mathbf{AbMon}$ as a subcategory of $\mathbf{Cat}$). I don't, however, know whether arbitrary presheaves $\Gamma^{\mathrm{op}} \to \mathbf{Set}_*$ get mapped to the same simplicial set under the different functors induced by different labellings.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.