### The Univalent Perspective on Classifying Spaces

#### Posted by Mike Shulman

I feel like I should apologize for not being more active at the Cafe recently. I’ve been busy, of course, and also most of my recent blog posts have been going to the HoTT blog, since I felt most of them were of interest only to the HoTT crowd (by which I mean, “people interested enough in HoTT to follow the HoTT blog” — which may of course include many Cafe readers as well). But today’s post, while also inspired by HoTT, is less technical and (I hope) of interest even to “classical” higher category theorists.

In general, a *classifying space for bundles of $X$’s* is a space $B$ such that maps $Y\to B$ are equivalent to bundles of $X$’s over $Y$. In classical algebraic topology, such spaces are generally constructed as the geometric realization of the nerve of a category of $X$’s, and as such they may be hard to visualize geometrically. However, it’s generally useful to think of $B$ as *a space whose points are $X$’s*, so that the classifying map $Y\to B$ of a bundle of $X$’s assigns to each $y\in Y$ the corresponding fiber (which is an $X$). For instance, the classifying space $B O$ of vector bundles can be thought of as a space whose points are vector spaces, where the classifying map of vector bundle assigns to each point the fiber over that point (which is a vector space).

In classical algebraic topology, this point of view can’t be taken quite literally, although we can make some use of it by identifying a classifying space with its representable functor. For instance, if we want to define a map $f:B O\to B O$, we’d like to say “a point $v\in B O$ is a vector space, so let’s do *blah* to it and get another vector space $f(v)\in B O$. We can’t do that, but we can do the next best thing: if *blah* is something that can be done fiberwise to a vector bundle in a natural way, then since $Hom(Y,B O)$ is naturally equivalent to the collection of vector bundles over $Y$, our *blah* defines a natural transformation $Hom(-,B O) \to Hom(-,B O)$, and hence a map $f:B O \to B O$ by the Yoneda lemma.

However, in higher category theory and homotopy type theory, we can really take this perspective literally. That is, if by “space” we choose to mean “$\infty$-groupoid” rather than “topological space up to homotopy”, then we can really define the classifying space to be *the $\infty$-groupoid of $X$’s*, whose points (objects) are $X$’s, whose morphisms are equivalences between $X$’s, and so on. Now, in defining a map such as our $f$, we can actually just give a map from $X$’s to $X$’s, as long as we check that it’s functorial on equivalences — and if we’re working in HoTT, we don’t even have to do the second part, since everything we can write down in HoTT is automatically functorial/natural.

This gives a different perspective on some classifying-space constructions that can be more illuminating than a classical one. Below the fold I’ll discuss some examples that have come to my attention recently.

All of these examples have to do with the classifying space of “types equivalent to $X$” for some fixed $X$. Such a classifying space, often denoted $B Aut(X)$, has the property that maps $Y \to B Aut(X)$ are equivalent to maps (perhaps “fibrations” or “bundles”) $Z\to Y$ all of whose fibers are equivalent (a homotopy type theorist might say “merely equivalent”) to $X$. The notation $B Aut(X)$ accords with the classical notation $B G$ for the delooping of a (perhaps $\infty$-) group: in fact this is a delooping of the group of automorphisms of $X$.

Categorically (and homotopy-type-theoretically), we simply define $B Aut(X)$ to be the *full* sub-$\infty$-groupoid of $\infty Gpd$ (the $\infty$-groupoid of $\infty$-groupoids) whose objects are those equivalent to $X$. You might have thought I was going to say the full sub-$\infty$-groupoid on the *single object* $X$, and that would indeed give us an *equivalent* result, but the examples I’m about to discuss really do rely on having all the other equivalent objects in there. In particular, note that an arbitrary object of $B Aut(X)$ is an $\infty$-groupoid that admits *some* equivalence to $X$, but *no such equivalence has been specified*.

### Example 1: $B Aut(2)$

As the first example, let $X = 2 = \{0,1\}$, the standard discrete space with two points. Then $Aut(2) = C_2$, the cyclic group on 2 elements, and so $B Aut(2) = B C_2 = K(C_2,1)$. Since $C_2$ is an abelian group, $B C_2$ again has a (2-)group structure, i.e. we should have a multiplication operation $B C_2 \times B C_2 \to B C_2$, an identity, inversion, etc.

Using the equivalence $B C_2 \simeq B Aut(2)$, we can describe all of these operations directly. A point $Z \in B Aut(2)$ is a space that’s equivalent to $2$, but without a specified equivalence. Thus, $Z$ is a set with two elements, but we haven’t chosen either of those elements to call “$0$” or “$1$”. As long as we perform constructions on $Z$ without making such an unnatural choice, we’ll get maps that act on $B Aut(2)$ and hence $B C_2$ as well.

The identity element of $B Aut(2)$ it’s fairly obvious: there’s only one canonical element of $B Aut(2)$, namely $2$ itself. The multiplication is not as obvious, and there may be more than one way to do it, but after messing around with it a bit you may come to the same conclusion I did: the product of $Z,W\in B Aut(2)$ should be $Iso(Z,W)$, the set of isomorphisms between $Z$ and $W$. Note that when $Z$ and $W$ are 2-element sets, so is $Iso(Z,W)$, but in general there’s no way to distinguish either of those isomorphisms from the other one, nor is $Iso(Z,W)$ naturally isomorphic to $Z$ or $W$. It is, however, obviously commutative: $Iso(Z,W) \cong Iso(W,Z)$.

Moreover, if $Z=2$ is the identity element, then $Iso(2,W)$ *is* naturally isomorphic to $W$: we can define $Iso(2,W) \to W$ by evaluating at $0\in 2$. Similarly, $Iso(Z,2)\cong Z$, so our “identity element” has the desired property.

Furthermore, if $Z=W$, then $Iso(Z,Z)$ *does* have a distinguished element, namely the identity. Thus, it naturally equivalent to $2$ by sending the identity to $0\in 2$. So every element of $B Aut(2)$ is its own inverse. The trickiest part is proving that this operation is associative. I’ll leave that to the reader (or you can try to decipher my Coq code).

(We did have to make some choices about whether to use $0\in 2$ or $1\in 2$. I expect that as long as we make those choices consistently, making them differently will result in equivalent 2-groups.)

### Example 2: An incoherent idempotent

In 1-category theory, an idempotent is a map $f:A\to A$ such that $f \circ f = f$. In higher category theory, the equality $f \circ f = f$ must be weakened to an isomorphism or equivalence, and then treated as extra data on which we ought to ask for additional axioms, such as that the two induced equivalences $f \circ f \circ f \simeq f$ coincide (up to an equivalence, of course, which then satisfies its own higher laws, etc.).

A natural question is if we have only an equivalence $f \circ f \simeq f$, whether it can be “improved” to a “fully coherent” idempotent in this sense. Jacob Lurie gave the following counterexample in Warning 1.2.4.8 of Higher Algebra:

let $G$ denote the group of homeomorphisms of the unit interval $[0,1]$ which fix the endpoints (which we regard as a discrete group), and let $\lambda : G \to G$ denote the group homomorphism given by the formula

$\lambda(g)(t) = \begin{cases} \frac{1}{2} g(2t) & \quad if\; 0\le t \le \frac{1}{2}\\ t & \quad if\; \frac{1}{2}\le t \le 1. \end{cases}$

Choose an element $h\in G$ such that $h(t)=2t$ for $0\le t\le \frac{1}{4}$. Then $\lambda(g)\circ h = h\circ \lambda(\lambda(g))$ for each $g\in G$, so that the group homomorphisms $\lambda,\lambda^2 : G\to G$ are conjugate to one another. It follows that the induced map of classifying spaces $e:B G \to B G$ is homotopic to $e^2$, and therefore idempotent in the homotopy category of spaces. However… $e$ cannot be lifted to a [coherent] idempotent in the $\infty$-category of spaces.

Let’s describe this map $e$ in the more direct way I suggested above. Actually, let’s do something easier and just as good: let’s replace $[0,1]$ by Cantor space $2^{\mathbb{N}}$. It’s reasonable to guess that this should work, since the essential property of $[0,1]$ being used in the above construction is that it can be decomposed into two pieces (namely $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$) which are both equivalent to itself, and $2^{\mathbb{N}}$ has this property as well:

$2^{\mathbb{N}} \cong 2^{\mathbb{N}+1} \cong 2^{\mathbb{N}} \times 2^1 \cong 2^{\mathbb{N}} + 2^{\mathbb{N}}.$

Moreover, $2^{\mathbb{N}}$ has the advantage that this decomposition is *disjoint*, i.e. a coproduct. Thus, we can also get rid of the assumption that our automorphisms preserve endpoints, which was just there in order to allow us to glue two different automorphisms on the two copies in the decomposition.

Therefore, our goal is now to construct an endomap of $B Aut(2^{\mathbb{N}})$ which is incoherently, but not coherently, idempotent. As discussed above, the elements of $B Aut(2^{\mathbb{N}})$ are spaces that are *equivalent* to $2^{\mathbb{N}}$, but without any such specified equivalence. Looking at the definition of Lurie’s $\lambda$, we can see that intuitively, what it does is shrink the interval to half of itself, acting functorially, and add a new copy of the interval at the end. Thus, it’s reasonable to define $e:B Aut(2^{\mathbb{N}}) \to B Aut(2^{\mathbb{N}})$ by

$e(Z) = Z + 2^{\mathbb{N}}.$

Here $Z$ is some space equivalent to $2^{\mathbb{N}}$, and in order for this map to be well-defined, we need to show is that if $Z$ is equivalent to $2^{\mathbb{N}}$, then so is $Z + 2^{\mathbb{N}}$. However, the decomposition $2^{\mathbb{N}} \cong 2^{\mathbb{N}} + 2^{\mathbb{N}}$ ensures this. Moreover, since our definition didn’t involve making any unnatural choices, it’s “obviously” (and in HoTT, automatically) functorial.

Now, is $e$ incoherently-idempotent, i.e. do we have $e(e(Z))\cong e(Z)$? Well, that is just asking whether

$(Z + 2^{\mathbb{N}}) + 2^{\mathbb{N}} \quad\text{is equivalent to}\quad Z + 2^{\mathbb{N}}$

but this again follows from $2^{\mathbb{N}} \cong 2^{\mathbb{N}} + 2^{\mathbb{N}}$! Showing that $e$ is not coherent is a bit harder, but still fairly straightforward using our description; I’ll leave it as an exercise, or you can try to decipher the Coq code.

### Example 3: Natural pointed sets

Let’s end by considering the following question: in what cases does the natural map $B S_{n-1} \to B S_{n}$ have a retraction, where $S_n$ is the symmetric group on $n$ elements? Looking at homotopy groups, this would imply that $S_{n-1} \hookrightarrow S_n$ has a retraction, which is true for $n\lt 5$ but not otherwise. But let’s look instead at the map on classifying spaces.

The obvious way to think about this map is to identify $B S_n$ with $B Aut(\mathbf{n})$, where $\mathbf{n}$ is the discrete set with $n$ elements, and similarly $B S_{n-1}$ with $B Aut(\mathbf{n-1})$. Then an element of $B Aut(\mathbf{n-1})$ is a set $Z$ with $n-1$ elements, and the map $B S_{n-1} \to B S_{n}$ takes it to $Z+1$ which has $n$ elements.

However, another possibility is to identify $B S_{n-1}$ instead with the classifying space of *pointed* sets with $n$ elements. Since an isomorphism of pointed sets must respect the basepoint, this gives an equivalent groupoid, and now the map $B S_{n-1} \to B S_{n}$ is just forgetting the basepoint. With this identification, a putative retraction $B S_{n} \to B S_{n-1}$ would assign, to any set $Z$ with $n$ elements, a *pointed* set $(r(Z),r_0)$ with $n$ elements. Note that the underlying set $r(Z)$ need not be $Z$ itself; they will of course be isomorphic (since both have $n$ elements), but there is no *specified* or natural isomorphism. However, to say that $r$ is a retraction of our given map says that *if* $Z$ started out pointed, *then* $(r(Z),r_0)$ is isomorphic to $(Z,z_0)$ as pointed sets.

Let’s do some small examples. When $n=1$, our map $r$ has to take a set with 1 element and assign to it a pointed set with 1 element. There’s obviously a unique way to do that, and just as obviously if we started out with a pointed set we get the same set back again.

The case $n=2$ is a bit more interesting: our map $r$ has to take a set $Z$ with 2 elements and assign to it a pointed set with 2 elements. One option, of course, is to define $r(Z)=2$ for all $Z$. Since every pointed 2-element set is uniquely isomorphic to every other, this satisfies the requirement. Another option motivated by example 1, which is perhaps a little more satisfying, would be to define $r(Z) = Iso(Z,Z)$, which is pointed by the identity.

The case $n=3$ is more interesting still, since now it is *not* true that any two pointed 3-element sets are naturally isomorphic. Given a 3-element set $Z$, how do we assign to it functorially a pointed 3-element set? The best way I’ve thought of is to let $r(Z)$ be the set of automorphisms $f\in Iso(Z,Z)$ such that $f^3 = id$. This has 3 elements, the identity and two 3-cycles, and we can take the identity as a basepoint. And if $Z$ came with a point $z_0$, then we can define an isomorphism $Z \cong r(Z)$ by sending $z\in Z$ to the unique $f\in r(Z)$ having the property that $f(z_0)= z$.

The case $n=4$ is somewhat similar: given a 4-element set $Z$, define $r(Z)$ to be the set of automorphisms $f\in Iso(Z,Z)$ such that $f^2 = id$ *and* whose set of fixed points is either empty or all of $Z$. This has 4 elements and is pointed by the identity; in fact, it is the permutation representation of the Klein four-group. And once again, if $Z$ came with a point $z_0$, we can define $Z \cong r(Z)$ by sending $z\in Z$ to the unique $f\in r(Z)$ such that $f(z_0)= z$.

I will end with a question that I don’t know the answer to: is there any way to see from this perspective on classifying spaces that such a retraction *doesn’t* exist in the case $n=5$?

## Re: The Univalent Perspective on Classifying Spaces

(typography: the later $C_n$s … did we mean $S_n$ or $𝔖_n$, perhaps? Of course we want to avoid Σ in the present context)