It is well-known that the free symmetric monoidal category on one object is the category $\mathbb{F}$ of finite sets and bijections. This is supposed to be the categorification of the monoid of natural numbers, and its algebraic $K$-theory is given by the sphere spectrum $\mathbb{S}$, the free symmetric monoidal $\infty$-groupoid with inverses.

Is there a natural description of the free symmetric monoidal $\infty$-groupoid (resp. $\infty$-category, $(\infty,\infty)$-category) on one object?

  • 9
    $\begingroup$ Note that there's no way to generate noninvertible $n$-morphisms for $n > 0$ in the free symmetric monoidal $(\infty,\infty)$-category on one object, so it should be the same as the free symmetric monoidal $\infty$-groupoid on one object. Treated as a space, the latter should be the space of finite subsets of $\mathbb{R}^\infty$. $\endgroup$
    – John Baez
    Jul 25, 2021 at 5:11
  • $\begingroup$ A detail: $\mathbb F$ (with disjoint union $\uplus$ as the monoidal product and a singleton $\ast$ as the object) is the free symmetric monoidal category on one commutatively monoidal object. (If you take, say, $\mathbb F$ with cartesian product $\times$ and an empty set $\varnothing$ instead, then there is no monoidal functor from $(\mathbb F,\uplus)$ to $(\mathbb F,\times)$ that maps $\ast$ to $\varnothing$. The unique function from $\varnothing$ to $\ast$ would have to be mapped to a function from $\ast$ to $\varnothing$, and there is none.) $\endgroup$ Jul 25, 2021 at 6:07
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    $\begingroup$ @TobyBartels Are you sure? Note that the question is defining $\mathbb{F}$ as the groupoid of finite sets, which of I'm not mistaken is the universal symmetric monoidal $\infty$-category on one object since it's the symmetric monoidal envelope of the $\infty$-operad $Triv^\otimes$ $\endgroup$ Jul 25, 2021 at 8:13
  • $\begingroup$ @Denis Nardin : You're right! Then the only problem is that the link in the question is to the wrong category. (It's understandable, since the nlab doesn't have a page dedicated to the groupoid, and it is discussed as a groupoid briefly there, but somebody should add the well-known fact to it.) $\endgroup$ Jul 26, 2021 at 12:28

1 Answer 1


Yes, it is the same as $\mathbb{F}$.

As John Baez points out, it is the same as the free symmetric monoidal $\infty$-groupoid on one object. (This can also be seen by playing around with the adjoints between groupoids and categories).

Symmetric monoidal $\infty$-groupoids are the same as $E_\infty$-spaces (careful: not assumed to be "grouplike"). So we want the free $E_\infty$-space on a point. As a space this is homotopy equivalent to $\sqcup_{n\geq0} B \Sigma_n$. Note all the hom spaces are actually 1-types, so it can be modeled as a 1-groupoid, namely $\mathbb{F}$. It might seem surprising at first that the free symmetric monoidal $\infty$-category is actually an ordinary category. However on reflection we see that it follows from the fact that the spaces in the $E_\infty$-operad are actually 1-types.

Equivalently, as John says, it is the space of all finite subsets of $\mathbb{R}^\infty$. If you pick an identification of $\mathbb{R}^\infty$ with the "interior" of an infinite cube, then you get a natural $E_\infty$ structure on this model of the space, too.

You can also see that it is $\mathbb{F}$ in the way that Denis suggests in his comment to the OP.

  • $\begingroup$ This is a really nice answer! Thank you! $\endgroup$
    – Emily
    Jul 30, 2021 at 3:42

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