TITLE: Are free $E_\infty$-algebras in spaces closed under pullbacks? (as a full subcategory of all $E_\infty$-algebras in spaces)
QUESTION [12 upvotes]: Let's call an $E_\infty$-algebra $A$ in spaces free if there is a space $A_0$ and an equivalence of $E_\infty$-algebras:
$
  \coprod_{n \ge 0} (A_0)^n_{h\Sigma_n} \simeq A.
$
Consider a diagram of $E_\infty$-algebras in spaces
$$
  A \longrightarrow B \longleftarrow C
$$
and assume that each of them is free in the above sense.
However, the maps $A \to B \leftarrow C$ are not required to be free, i.e. they don't have to send $A_0$ to $B_0$ or $C_0$ to $B_0$.

In this setting, is the homotopy pullback $A \times_B C$ always free?


I've convinced myself in a very roundabout way that this should be true, but would be very grateful for a proper proof or even better a reference.
I think a hands-on approach using the formula for the free $E_\infty$-algebra might work, but the combinatorics will get quite tricky.
As an example, consider the case where $A, B, C$ are all free on a point and the maps $A \to B \leftarrow C$ both send the generator to twice the generator.
Then I think the resulting pullback $A \times_B C$ is free on a space with infinitely many components.
Let me give some more details on what happens in this example. I'll think of all the involved 1-types as groupoids, so $A, B, C$ are all represented by the groupoid of finite sets and bijections $\mathrm{FB}$, which is symmetric monoidal under disjoint union.
In my example the two maps correspond to the functor $F: \mathrm{FB} \to \mathrm{FB}$ that sends $a \mapsto a \times \{0,1\}$.
Now the homotopy pullback can be computed in terms of groupoids. The resulting groupoid $\mathcal{G}$ has as objects triples $(a, b, \varphi)$ where $a, b \in \mathrm{FB}$ are finite sets and $\varphi: a \times \{0,1\} \cong b \times \{0,1\}$ is a bijection. Morphisms are tuples of bijections compatible with $\varphi$.
Define the set of connected components of such an object $(a, b, \varphi)$ as the pushout $\pi_0(a,b,\varphi) = a \amalg_{a \times \{0,1\}} b$ where the right-hand map is $\varphi$ composed with projection to $b$.
This defines a symmetric monoidal functor $\pi_0: \mathcal{G} \to \mathrm{FB}$.
Let's say an object is connected if $\pi_0(a,b,\varphi)$ is a point.
Then I claim that $\mathcal{G}$ is freely generated under disjoint union by connected objects.
This is probably easiest to see by thinking of the objects as directed graphs with red and blue edges, where the vertices are $a \times \{0,1\}$, the blue edges are $(a, 0) \to (a, 1)$, the red edges are $\varphi^{-1}(b,0) \to \varphi^{-1}(b,1)$. Only graphs where every vertex is incident to exactly one blue and one red edge are allowed.
Now this category is a free symmetric monoidal groupoid on the connected graphs, of which there are countably infinitely many.

REPLY [4 votes]: This is not an answer, but I think it's really interesting to look at an example where this statement fails for commutative monoids and the analogous statement for $E_\infty$-algebras works out. This is too long for a comment, so here it goes:
By reverse engeneering the theorem about Krull monoids that @TimCampion mentions in his comment, we can find the following minimal example.
Consider the pullback of free commutative monoids:
$$
 \mathbb{N}^2 \xrightarrow{\;\;+\;\;} \mathbb{N} \xleftarrow{\;\;\cdot2\;\;} \mathbb{N}
$$
The pullback is $\{(x,y) \in \mathbb{N}^2 | x + y \equiv 0 \mod 2\}$. This can't be a free commutative monoid: any generating set must contain $\{(2,0), (1,1), (0,2)\}$ (and this is a generating set), so it would have to be free on at least three generators. On the other hand we know that it's group-completion is $\mathbb{Z}^2$, so if it was free it would have to have two generators.
Why does this not give a counter-example to the $E_\infty$-algebra question?
Let's model the free $E_\infty$-algebra on a point by $\mathrm{FB}$, the groupoid of finite sets and bijections, and accordingly the free $E_\infty$-algebra on two points by $\mathrm{FB}^2$, the groupoid of pairs of finite sets.
Then the analogous diagram to the above would be:
$$
 \mathrm{FB}^2 \xrightarrow{\;\;\amalg\;\;} \mathrm{FB} \xleftarrow{\;\;\times \{1,2\}\;\;} \mathrm{FB}
$$
An object in the homotopy pullback groupoid can be described as a triple $(X, Y, Z)$ of finite sets together with a bijection $\varphi:X \amalg Y \cong Z \times \{1,2\}$.
We can forget about $\varphi$ and $Z$ and instead remember the directed pairing induced by pairing $\varphi^{-1}(z,1)$ with $\varphi^{-1}(z,2)$.
So the pullback is the groupoid of tuples $(X, Y)$ together with a directed pairing on $X \amalg Y$.
This is symmetric monoidal under disjoint union and as a symmetric monoidal groupoid it's generated by four objects:
$(\{a,b\}, \emptyset)$, $(\{a\}, \{b\})$, $(\{b\}, \{a\})$ and $(\emptyset, \{a,b\})$ each with the directed pairing $a \to b$.
So in this $E_\infty$-algebra setting the pullback is actually free on four generators.