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.

Krull monoidis a finite limit of of maps between finitely-generated free commutative monoids -- (or equivalently, the set of nonnegative integer solutions to some integer equations in $\mathbb Z^m$ -- see Thm 8.7 in Grillet'sCommutative Semigroups). Just going by the name, Ithinkthere exist Krull monoids which are not free. So free commutative monoids arenotclosed under finite limits. Of course, as you say, the situation may be quite different in spaces. $\endgroup$7more comments