# Locally constant (homotopy) pre-factorization algebras

In my thesis, I'm using the theory of (homotopy) factorization algebras and particularly locally constant ones. While reading an article that I can't find again I read that already a locally constant pre-factorization algebra $$F$$ is automatically a homotopy factorization algebra. For me locally constant means that when $$U\subseteq V$$ are homeomorphic to disks in $$\mathbb R^n$$, then $$F(U)\rightarrow F(V)$$ is a quasi-isomorphism of chain complexes.

The base manifold $$M$$ is a smooth manifold in the case I'm interested in.

My motivation (which is far from an argument) is that we can refine a Weiss (or factorizing) open cover of $$U$$ to a good Weiss open cover $$\mathcal W$$ of by breaking each element in the Weiss cover into disks (i.e. subsets homeomorphic to disks). Then the homotopy cosheaf condition became that $$F(U)$$ is quasi-isomorphic to the homotopy colimit of a constant diagram, where the $$n$$-th object is $$\coprod_{U_1,\cdots,U_n\in\mathcal W}F(U_1\cap\cdots\cap U_n)$$ and $$U_1\cap \cdots\cap U_n$$ are disks.

As I said, my motivation is very far from an actual argument, so if someone has a real proof (or counterexample or a reference), that would be great.

Edit: Ok, the comment by Dmitri made (nice counterexample too) me realize that I haven't specified that the pre-factorization algebras that I'm working with are lax monoidal in the sense that given disjoint open subsets $$U_1,\cdots,U_n$$, the map $$F(U_1)\otimes\cdots\otimes F(U_n)\rightarrow F(U_1\sqcup\cdots\sqcup U_n)$$ is an isomorphism. So my follow-up question is: Does this new detail affect the validity of the claim? As Dmitri points out, without the above extra condition, the claim is definitely not true.

• For a simple counterexample, consider the case when M is the disjoint union of two points. Then the local constancy condition is always true. A prefactorization algebra on M is two objects A, B assigned to the two connected components, together with a map A⊗B→C. For a factorization algebra, the latter map must be a weak equivalence, which is a stronger condition. Jun 15 at 21:18
• @DmitriPavlov, thank you for the counterexample. I've edited the question with additional detail about the types of pre-factorization algebras I'm working with, if you know a counterexample for this new situation as well, I'd love to hear it. Jun 15 at 21:45
• For a counterexample for the updated statement, consider the case when M is the disjoint union of countably many points. As before, local constancy trivially holds. A prefactorization algebra on M is a family of objects A_i together with pointings 1→A_i, as well as a map ⨂_i A_i → B, where B is the value on M and ⨂ denotes the infinite monoidal product that uses the pointings of A_i. A factorization algebra further requires that the latter map is a weak equivalence, which is a stronger condition. Jun 15 at 22:50
• I see, so the statement seems to be just false. Good to know, thank a lot @DmitriPavlov Jun 15 at 22:59
• In your updated statement the symmetric monoidal condition allows one to define an excisivity condition for unions of submanifolds. Perhaps the statement you are referring to is saying that a locally constant homotopy prefactorization algebra which satisfies excision is a factorization algebra? Jun 16 at 4:27