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.

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